{"id":568,"date":"2021-07-21T14:21:05","date_gmt":"2021-07-21T18:21:05","guid":{"rendered":"http:\/\/pressbooks.library.upei.ca\/statics\/?post_type=chapter&#038;p=568"},"modified":"2025-07-31T22:31:27","modified_gmt":"2025-08-01T02:31:27","slug":"2-4-examples","status":"publish","type":"chapter","link":"https:\/\/pressbooks.library.upei.ca\/statics\/chapter\/2-4-examples\/","title":{"raw":"2.4. Examples","rendered":"2.4. Examples"},"content":{"raw":"Here are examples from Chapter 2 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 2.4.1: Equilibrium Equation and Components of Vectors, Submitted by Analiya Benny<\/h1>\r\n<div class=\"textbox shaded\">\r\n<ol>\r\n \t<li><strong>Problem<\/strong><\/li>\r\n<\/ol>\r\nA 280 lb Pipe is being lifted by a crane as shown in the figure. Find the magnitude of the tension on the cables, given that the forces on the cables are equal.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/image-chap-2-PB-pipe-e1667658386342.jpg\" alt=\"A cylindrical steel pipe suspended by two cables and a crane hook.\" width=\"462\" height=\"451\" class=\"aligncenter wp-image-1962 size-full\" \/>\r\n\r\nhttps:\/\/publicdomainvectors.org\/en\/free-clipart\/Crane-with-a-pipeai\/86375.html\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-2025-05-02-125158-300x254.png\" alt=\"A sketch of the problem\" width=\"312\" height=\"264\" class=\"aligncenter wp-image-2281\" \/>\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<strong>2. Draw<\/strong>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-6-e1667699434378.png\" alt=\"A free-body diagram of the problem\" width=\"492\" height=\"480\" class=\"alignnone wp-image-1967 size-full\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>F = 280 lb<\/li>\r\n \t<li>\u03b8 = 60<span>\u00b0\u00a0<\/span><\/li>\r\n<\/ul>\r\nUnknown:\r\n<ul>\r\n \t<li>[latex]\\overrightarrow{T_{B}}[\/latex]<\/li>\r\n \t<li>[latex]\\overrightarrow{T_{A}}[\/latex]<\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nApplying the equilibrium equation and componentization of vectors to find [latex]T_{B}[\/latex] and [latex]T_{A}[\/latex].\r\n\r\n<strong>Note:<\/strong> [latex]T_{B}[\/latex] = [latex]T_{A}[\/latex]\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\n<span style=\"text-decoration: underline\">Components of [latex]T_{B}[\/latex]<\/span>\r\n\r\n[latex]T_{By} = T_{B} \\sin {60}[\/latex]\r\n\r\nand, [latex]T_{Bx} = T_{B}\\cos {60}[\/latex]\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/2.4.1done-300x192.jpg\" alt=\"A force vector T_B at a 60\u00b0 angle resolved into horizontal T_Bx and vertical T_By components.\" width=\"300\" height=\"192\" class=\"alignnone wp-image-2089 size-medium\" \/>\r\n\r\n<span style=\"text-decoration: underline\">Equilibrium equation in the y direction<\/span>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-8-e1667700978863.png\" alt=\"A 280 lb load suspended by two symmetric cables at 60\u00b0 angles with equal tension forces T_B. \" width=\"544\" height=\"408\" class=\"alignnone wp-image-1970 size-full\" \/>\r\n\r\n[latex]\\sum F_{y}= -T_{By} -T_{By}-280 lb = 0 \\\\\r\n\r\nT_{By} = \\dfrac{-280 lb}{2} = -140 lb [\/latex]\r\n\r\nNote that the -ve sign here means that the direction of vector T<sub>B<\/sub> is opposite to that drawn in the FBD diagram. So now the FBD is as given below:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-10-e1667702644190.png\" alt=\"A 280 lb load suspended by two upward-angled cables at 60\u00b0, each exerting a tension force T_B.\" width=\"464\" height=\"420\" class=\"alignnone wp-image-1975 size-full\" \/>\r\n\r\nnow,\u00a0 [latex]T_{B} \\sin {60} = 140 lb [\/latex]\r\n\r\n[latex]T_{B} = \\dfrac{140 lb}{\\sin{60}}[\/latex]\r\n\r\nThus, [latex]T_{B} = T_{A} = 161.66 lb[\/latex]\r\n\r\nThe magnitude of the tension carried by each of the cables is 161.66 lb.\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nThe results show that the x component does no essential contribution towards carrying the weight of the pipe because the sum of forces in the x direction just cancels out and equals 0lb.\r\n\r\n[latex]\\sum F_{x}= -T_{Bx} +T_{Bx}= 0[\/latex]\r\n<h1>Example 2.4.2: Equilibrium Equation, Submitted by Kylian Duplan<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nDesmond (80kg) is hanging for his life in a well, on a horizontal pole that is between two brick walls that displace 1200N of force each upwards along the pole. There is also a single brick positioned 2m away from the first wall. If Desmond is hanging 3m away from the second wall, and is to catch a backpack falling above him, what mass should the backpack be to break the pole, causing Desmond to fall into the well? What is the mass of the single brick? [g=9.81m\/s^2]\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture8-300x279.jpg\" alt=\"An 80 kg person hanging from a horizontal bar stretched between two walls, with distances and directions labeled.\" width=\"442\" height=\"411\" class=\"alignnone wp-image-2572\" \/>\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<strong>2. Draw\u00a0<\/strong>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-171126-300x173.png\" alt=\"A free-body diagram of the problem\" width=\"447\" height=\"258\" class=\"alignnone wp-image-2574\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>Desmond\u2019s mass = 80kg<\/li>\r\n \t<li>Mass of bag caught = (y)kg<\/li>\r\n \t<li>Weight of Desmond + bag, W = 9.81(80 + y)<\/li>\r\n \t<li>Weight of the block = x<\/li>\r\n \t<li>Wall A = First wall<\/li>\r\n \t<li>Wall B = Second wall<\/li>\r\n \t<li>g = 9.81m\/s<sup>2<\/sup><\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>x<\/li>\r\n \t<li>y<\/li>\r\n<\/ul>\r\n<strong>4. Approach\u00a0<\/strong>\r\n\r\nApply equilibrium equations and solve simultaneously to find x and y\r\n\r\n<strong>5. Analysis\u00a0<\/strong>\r\n\r\nTaking moment about wall B,\r\n<div>[latex]\\sum M_B = W(3\\,\\mathrm{m}) + x(10 - 2)\\,\\mathrm{m} - 1200(10\\,\\mathrm{m}) = 0[\/latex]<\/div>\r\n<div><\/div>\r\n<div>[latex]3 \\times 9.81(80 + y) + 8x - 12000 = 0[\/latex]<\/div>\r\n<div><\/div>\r\n<div>[latex]29.43y + 8x = 9645.6 \\quad \\text{(1)}[\/latex]<\/div>\r\n<div>\r\n\r\nSumming up vertical forces\r\n<div>[latex]\\sum F_y = 1200 + 1200 - x - W = 0[\/latex]<\/div>\r\n<div><\/div>\r\n<div>[latex]2400 - x - 9.81(80 + y) = 0[\/latex]<\/div>\r\n<div><\/div>\r\n<div>[latex]2400 - x - 784.8 - 9.81y = 0[\/latex]<\/div>\r\n<div><\/div>\r\n<div>[latex]x + 9.81y = 1615.2 \\quad \\text{(2)}[\/latex]<\/div>\r\n<div><\/div>\r\n<div>\r\n<div>Solving simultaneously;<\/div>\r\n<div>[latex]8x + 29.43y = 9645.6 \\quad \\text{(1)}[\/latex]<\/div>\r\n<div><\/div>\r\n<div>[latex]x + 9.81y = 1615.2 \\quad \\text{(2)}[\/latex]<\/div>\r\n<div><\/div>\r\n<div>[latex]\\Rightarrow x = 960\\,\\mathrm{N}, \\quad y = 66.79\\,\\mathrm{N}[\/latex]<\/div>\r\n<div>\r\n<div>\r\n\r\nIn kilograms; (Divide by 9.81m\/s<sup>2<\/sup>)\r\n\r\n<\/div>\r\n<div>[latex]x = \\frac{960}{9.81} = 97.86\\,\\mathrm{kg}, \\quad y = \\frac{66.79}{9.81} = 6.81\\,\\mathrm{kg}[\/latex]<\/div>\r\n<div><\/div>\r\n<\/div>\r\n<\/div>\r\n<strong style=\"text-align: initial;background-color: initial;font-size: 1em\">6. Review\u00a0<\/strong>\r\n\r\nThe block would need to weigh 97.86 kg at 2m from Wall A, and Desmond should catch the bag weighing 6.81kg while he hangs 7m from Wall A.\r\n\r\n<\/div>\r\n<\/div>\r\n<h1>Example 2.4.3: Equilibrium Equation, Submitted Anonymously<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nA military M35 truck is lifted using a helicopter hover lift and cables whose upper strings are slightly shorter than the lower strings. The shorter-upper strings create an angle of 40 degrees to the top of the truck, and the longer-lower strings create an angle of 60 degrees to the top of the truck.\u00a0 If the truck has a mass of 5000 kg and is evenly distributed, determine:\r\n\r\na. Draw Free Body Diagram at point A, B, and C.\r\n\r\nb. what are the tension on each string?\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/the-ch-53k-king-stallion-lifts-a-joint-light-tactical-4c867f-1024-300x200.jpg\" alt=\"A helicopter flying in the sky while carrying a military vehicle suspended by cables.\" width=\"300\" height=\"200\" class=\"alignnone wp-image-2107 size-medium\" \/>\r\n\r\nhttps:\/\/garystockbridge617.getarchive.net\/media\/the-ch-53k-king-stallion-lifts-a-joint-light-tactical-4c867f\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\/2021\/11\/helicopter-191x300.jpg\" alt=\"A digital sketch of the problem\" width=\"191\" height=\"300\" class=\"alignnone wp-image-2109 size-medium\" \/>\r\n\r\nFree Body Diagram:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/helicopterFBD2.jpg\" alt=\"A free-body diagram of the problem\" width=\"201\" height=\"251\" class=\"alignnone wp-image-2112 size-full\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>m=5000 kg<\/li>\r\n \t<li><span>\u2220A = 40\u00b0<\/span><\/li>\r\n \t<li><span>\u2220B = 60\u00b0<\/span><\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>F<sub>1<\/sub><\/li>\r\n \t<li>F<sub>2\u00a0<\/sub><\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nConsidering the load is evenly distributed, the upper pair of string have equal lengths, and so does the lower pair, FBD can be drawn for each points. Use equilibrium equations for x and y direction to find tension on each string.\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\nWeight,[latex]w =m\\times g= 5000 kg\\times 9.81= 49050 N[\/latex]\r\n\r\n<span style=\"text-decoration: underline\">Point A<\/span>\r\n\r\n[latex]\\begin{equation} \\sum{F_{y} = F_{1} \\sin{40} -\\dfrac{w}{4}} \\end {equation}[\/latex]\r\n\r\n[latex]\\begin{equation} F_{1} = \\dfrac{w}{4\\times \\sin{40}} = \\dfrac{49050}{4\\times 0.643} \\end{equation}[\/latex]\r\n\r\n[latex]F_{1} = 19070.76 N[\/latex]\r\n\r\n<span style=\"text-decoration: underline\">Point B<\/span>\r\n\r\n[latex]\\begin{equation} \\sum{F_{y} = F_{2} \\sin{60} -\\dfrac{w}{4}} \\end {equation}[\/latex]\r\n\r\n[latex]\\begin{equation} F_{2} = \\dfrac{w}{4\\times \\sin{60}} \\end{equation}[\/latex]\r\n\r\n[latex]F_{2} = 14159.51 N[\/latex]\r\n\r\n<span style=\"text-decoration: underline\">Point C<\/span>\r\n\r\n[latex]F_{T} =F_{1}\\sin{40} +F_{2}\\sin{60}[\/latex]\r\n\r\n[latex]\\begin{equation} \\quad\\quad\u00a0 = (19070.76\\times 0.643) +(14159.51\\times 0.866)\\end{equation}[\/latex]\r\n\r\n[latex]F_{T} = 24524.63 N[\/latex]\r\n\r\nThus both the upper strings are at a tension of 19070.76 N and lower strings are at a tension of 14159.51 N.\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nNote that in this problem the vehicle's weight was evenly distributed which is convenient in symmetrical situations like this. The tension on the strings can be verified as follows.\r\n\r\n[latex]2F_{T} = m\\times g [\/latex]\r\n\r\n<\/div>\r\n<h1>Example 2.4.4: Equilibrium Equation, Submitted by Analiya Benny<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nA child with a mass of 20kg is sitting on a swing with both strings 4 m long.\r\n\r\na. Draw the free-body diagram.\r\n\r\nb. Calculate the tension of the strings.\r\n\r\nc. What happens to the tension of the strings when string A is 3 m and string B is 4 m? Explain with a free-body diagram.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/child-girl-doll-happy-preview.jpg\" alt=\"Girl on a swing.\" width=\"499\" height=\"749\" class=\"alignnone wp-image-2059\" \/>\r\n\r\nImage 1:\u00a0 A real-life demonstration of the situation in the question.\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<strong>2. Draw\u00a0<\/strong>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-33-e1675991020826.png\" alt=\"A free-body diagram of the problem\" width=\"908\" height=\"468\" class=\"alignnone wp-image-2063 size-full\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>[latex]M= 20kg[\/latex]<\/li>\r\n \t<li>[latex]g = 9.81 m\/s^2[\/latex]<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>F<sub>A<\/sub><\/li>\r\n \t<li>F<sub>B<\/sub><\/li>\r\n<\/ul>\r\n<strong>4. Approach\u00a0<\/strong>\r\n\r\nAnalyzing the dimensions of the situation and applying the equilibrium equation.\r\n\r\n<strong>5. Analysis\u00a0<\/strong>\r\n\r\n[latex]w = m \\times g = 20 kg \\times 9.81= 196.2 N\u00a0 [\/latex]\r\n\r\nb. When the length of the strings is equal, the weight on point A is equal to the weight on point B. F<sub>A<\/sub> = F<sub>B<\/sub>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-31-e1675989644427.png\" alt=\"A suspended cable system with endpoints A and B, 4 meters apart vertically, showing forces F_A and F_B each supporting half the total weight w.\" width=\"708\" height=\"420\" class=\"alignnone wp-image-2061 size-full\" \/>\r\n\r\nNo Force in the x-direction\r\n\r\n[latex]\r\n\\sum F_y = 0 \\\\\r\nF_A + F_B = 196.2 \\, \\text{N} \\\\\r\nF_B + F_B = 196.2 \\\\\r\n2F_B = 196.2 \\\\\r\nF_B = 98.1 \\, \\text{N} \\\\\r\n\r\n\\text{Tension in each string is } 98.1 \\, \\text{N}\r\n[\/latex]\r\n\r\nc. When one chain of the swing is longer, the tension on the chains varies such that [latex]w_{B}&gt;w_{A}[\/latex]; in turn, chain B will experience more tension than chain A. To solve for tension in the chains, we would require more information about the distribution of the weights on points A and B.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-32-e1675990212710.png\" alt=\"An asymmetric cable system with endpoints A and B at different heights, showing unequal weight distribution where w_B' &gt; w_A.\" width=\"816\" height=\"448\" class=\"alignnone wp-image-2062 size-full\" \/>\r\n\r\n<strong>6. Review\u00a0<\/strong>\r\n\r\nIn questions involving free-body diagrams, it is convenient to have perfectly symmetrical situations. However, when faced with non-symmetrical or distributed loads, different techniques need to be applied, which will be explained in the following chapters.\r\n\r\n<\/div>\r\n&nbsp;","rendered":"<p>Here are examples from Chapter 2 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 2.4.1: Equilibrium Equation and Components of Vectors, Submitted by Analiya Benny<\/h1>\n<div class=\"textbox shaded\">\n<ol>\n<li><strong>Problem<\/strong><\/li>\n<\/ol>\n<p>A 280 lb Pipe is being lifted by a crane as shown in the figure. Find the magnitude of the tension on the cables, given that the forces on the cables are equal.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/image-chap-2-PB-pipe-e1667658386342.jpg\" alt=\"A cylindrical steel pipe suspended by two cables and a crane hook.\" width=\"462\" height=\"451\" class=\"aligncenter wp-image-1962 size-full\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/image-chap-2-PB-pipe-e1667658386342.jpg 462w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/image-chap-2-PB-pipe-e1667658386342-300x293.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/image-chap-2-PB-pipe-e1667658386342-65x63.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/image-chap-2-PB-pipe-e1667658386342-225x220.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/image-chap-2-PB-pipe-e1667658386342-350x342.jpg 350w\" sizes=\"auto, (max-width: 462px) 100vw, 462px\" \/><\/p>\n<p>https:\/\/publicdomainvectors.org\/en\/free-clipart\/Crane-with-a-pipeai\/86375.html<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-2025-05-02-125158-300x254.png\" alt=\"A sketch of the problem\" width=\"312\" height=\"264\" class=\"aligncenter wp-image-2281\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-2025-05-02-125158-300x254.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-2025-05-02-125158-65x55.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-2025-05-02-125158-225x191.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-2025-05-02-125158-350x297.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-2025-05-02-125158.png 551w\" sizes=\"auto, (max-width: 312px) 100vw, 312px\" \/><\/p>\n<\/div>\n<p>&nbsp;<\/p>\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\/11\/Screenshot-6-e1667699434378.png\" alt=\"A free-body diagram of the problem\" width=\"492\" height=\"480\" class=\"alignnone wp-image-1967 size-full\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-6-e1667699434378.png 492w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-6-e1667699434378-300x293.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-6-e1667699434378-65x63.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-6-e1667699434378-225x220.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-6-e1667699434378-350x341.png 350w\" sizes=\"auto, (max-width: 492px) 100vw, 492px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>F = 280 lb<\/li>\n<li>\u03b8 = 60<span>\u00b0\u00a0<\/span><\/li>\n<\/ul>\n<p>Unknown:<\/p>\n<ul>\n<li>[latex]\\overrightarrow{T_{B}}[\/latex]<\/li>\n<li>[latex]\\overrightarrow{T_{A}}[\/latex]<\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Applying the equilibrium equation and componentization of vectors to find [latex]T_{B}[\/latex] and [latex]T_{A}[\/latex].<\/p>\n<p><strong>Note:<\/strong> [latex]T_{B}[\/latex] = [latex]T_{A}[\/latex]<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p><span style=\"text-decoration: underline\">Components of [latex]T_{B}[\/latex]<\/span><\/p>\n<p>[latex]T_{By} = T_{B} \\sin {60}[\/latex]<\/p>\n<p>and, [latex]T_{Bx} = T_{B}\\cos {60}[\/latex]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/2.4.1done-300x192.jpg\" alt=\"A force vector T_B at a 60\u00b0 angle resolved into horizontal T_Bx and vertical T_By components.\" width=\"300\" height=\"192\" class=\"alignnone wp-image-2089 size-medium\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/2.4.1done-300x192.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/2.4.1done-65x42.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/2.4.1done-225x144.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/2.4.1done-350x224.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/2.4.1done.jpg 361w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p><span style=\"text-decoration: underline\">Equilibrium equation in the y direction<\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-8-e1667700978863.png\" alt=\"A 280 lb load suspended by two symmetric cables at 60\u00b0 angles with equal tension forces T_B.\" width=\"544\" height=\"408\" class=\"alignnone wp-image-1970 size-full\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-8-e1667700978863.png 544w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-8-e1667700978863-300x225.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-8-e1667700978863-65x49.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-8-e1667700978863-225x169.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-8-e1667700978863-350x263.png 350w\" sizes=\"auto, (max-width: 544px) 100vw, 544px\" \/><\/p>\n<p>[latex]\\sum F_{y}= -T_{By} -T_{By}-280 lb = 0 \\\\    T_{By} = \\dfrac{-280 lb}{2} = -140 lb[\/latex]<\/p>\n<p>Note that the -ve sign here means that the direction of vector T<sub>B<\/sub> is opposite to that drawn in the FBD diagram. So now the FBD is as given below:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-10-e1667702644190.png\" alt=\"A 280 lb load suspended by two upward-angled cables at 60\u00b0, each exerting a tension force T_B.\" width=\"464\" height=\"420\" class=\"alignnone wp-image-1975 size-full\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-10-e1667702644190.png 464w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-10-e1667702644190-300x272.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-10-e1667702644190-65x59.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-10-e1667702644190-225x204.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-10-e1667702644190-350x317.png 350w\" sizes=\"auto, (max-width: 464px) 100vw, 464px\" \/><\/p>\n<p>now,\u00a0 [latex]T_{B} \\sin {60} = 140 lb[\/latex]<\/p>\n<p>[latex]T_{B} = \\dfrac{140 lb}{\\sin{60}}[\/latex]<\/p>\n<p>Thus, [latex]T_{B} = T_{A} = 161.66 lb[\/latex]<\/p>\n<p>The magnitude of the tension carried by each of the cables is 161.66 lb.<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>The results show that the x component does no essential contribution towards carrying the weight of the pipe because the sum of forces in the x direction just cancels out and equals 0lb.<\/p>\n<p>[latex]\\sum F_{x}= -T_{Bx} +T_{Bx}= 0[\/latex]<\/p>\n<h1>Example 2.4.2: Equilibrium Equation, Submitted by Kylian Duplan<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>Desmond (80kg) is hanging for his life in a well, on a horizontal pole that is between two brick walls that displace 1200N of force each upwards along the pole. There is also a single brick positioned 2m away from the first wall. If Desmond is hanging 3m away from the second wall, and is to catch a backpack falling above him, what mass should the backpack be to break the pole, causing Desmond to fall into the well? What is the mass of the single brick? [g=9.81m\/s^2]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture8-300x279.jpg\" alt=\"An 80 kg person hanging from a horizontal bar stretched between two walls, with distances and directions labeled.\" width=\"442\" height=\"411\" class=\"alignnone wp-image-2572\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture8-300x279.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture8-65x60.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture8-225x209.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture8-350x325.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Picture8.jpg 750w\" sizes=\"auto, (max-width: 442px) 100vw, 442px\" \/><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p><strong>2. Draw\u00a0<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-171126-300x173.png\" alt=\"A free-body diagram of the problem\" width=\"447\" height=\"258\" class=\"alignnone wp-image-2574\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-171126-300x173.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-171126-65x37.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-171126-225x130.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-171126-350x201.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Screenshot-2025-07-09-171126.png 674w\" sizes=\"auto, (max-width: 447px) 100vw, 447px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>Desmond\u2019s mass = 80kg<\/li>\n<li>Mass of bag caught = (y)kg<\/li>\n<li>Weight of Desmond + bag, W = 9.81(80 + y)<\/li>\n<li>Weight of the block = x<\/li>\n<li>Wall A = First wall<\/li>\n<li>Wall B = Second wall<\/li>\n<li>g = 9.81m\/s<sup>2<\/sup><\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>x<\/li>\n<li>y<\/li>\n<\/ul>\n<p><strong>4. Approach\u00a0<\/strong><\/p>\n<p>Apply equilibrium equations and solve simultaneously to find x and y<\/p>\n<p><strong>5. Analysis\u00a0<\/strong><\/p>\n<p>Taking moment about wall B,<\/p>\n<div>[latex]\\sum M_B = W(3\\,\\mathrm{m}) + x(10 - 2)\\,\\mathrm{m} - 1200(10\\,\\mathrm{m}) = 0[\/latex]<\/div>\n<div><\/div>\n<div>[latex]3 \\times 9.81(80 + y) + 8x - 12000 = 0[\/latex]<\/div>\n<div><\/div>\n<div>[latex]29.43y + 8x = 9645.6 \\quad \\text{(1)}[\/latex]<\/div>\n<div>\n<p>Summing up vertical forces<\/p>\n<div>[latex]\\sum F_y = 1200 + 1200 - x - W = 0[\/latex]<\/div>\n<div><\/div>\n<div>[latex]2400 - x - 9.81(80 + y) = 0[\/latex]<\/div>\n<div><\/div>\n<div>[latex]2400 - x - 784.8 - 9.81y = 0[\/latex]<\/div>\n<div><\/div>\n<div>[latex]x + 9.81y = 1615.2 \\quad \\text{(2)}[\/latex]<\/div>\n<div><\/div>\n<div>\n<div>Solving simultaneously;<\/div>\n<div>[latex]8x + 29.43y = 9645.6 \\quad \\text{(1)}[\/latex]<\/div>\n<div><\/div>\n<div>[latex]x + 9.81y = 1615.2 \\quad \\text{(2)}[\/latex]<\/div>\n<div><\/div>\n<div>[latex]\\Rightarrow x = 960\\,\\mathrm{N}, \\quad y = 66.79\\,\\mathrm{N}[\/latex]<\/div>\n<div>\n<div>\n<p>In kilograms; (Divide by 9.81m\/s<sup>2<\/sup>)<\/p>\n<\/div>\n<div>[latex]x = \\frac{960}{9.81} = 97.86\\,\\mathrm{kg}, \\quad y = \\frac{66.79}{9.81} = 6.81\\,\\mathrm{kg}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<p><strong style=\"text-align: initial;background-color: initial;font-size: 1em\">6. Review\u00a0<\/strong><\/p>\n<p>The block would need to weigh 97.86 kg at 2m from Wall A, and Desmond should catch the bag weighing 6.81kg while he hangs 7m from Wall A.<\/p>\n<\/div>\n<\/div>\n<h1>Example 2.4.3: Equilibrium Equation, Submitted Anonymously<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>A military M35 truck is lifted using a helicopter hover lift and cables whose upper strings are slightly shorter than the lower strings. The shorter-upper strings create an angle of 40 degrees to the top of the truck, and the longer-lower strings create an angle of 60 degrees to the top of the truck.\u00a0 If the truck has a mass of 5000 kg and is evenly distributed, determine:<\/p>\n<p>a. Draw Free Body Diagram at point A, B, and C.<\/p>\n<p>b. what are the tension on each string?<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/the-ch-53k-king-stallion-lifts-a-joint-light-tactical-4c867f-1024-300x200.jpg\" alt=\"A helicopter flying in the sky while carrying a military vehicle suspended by cables.\" width=\"300\" height=\"200\" class=\"alignnone wp-image-2107 size-medium\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/the-ch-53k-king-stallion-lifts-a-joint-light-tactical-4c867f-1024-300x200.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/the-ch-53k-king-stallion-lifts-a-joint-light-tactical-4c867f-1024-768x512.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/the-ch-53k-king-stallion-lifts-a-joint-light-tactical-4c867f-1024-65x43.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/the-ch-53k-king-stallion-lifts-a-joint-light-tactical-4c867f-1024-225x150.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/the-ch-53k-king-stallion-lifts-a-joint-light-tactical-4c867f-1024-350x233.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/the-ch-53k-king-stallion-lifts-a-joint-light-tactical-4c867f-1024.jpg 1024w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>https:\/\/garystockbridge617.getarchive.net\/media\/the-ch-53k-king-stallion-lifts-a-joint-light-tactical-4c867f<\/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\/2021\/11\/helicopter-191x300.jpg\" alt=\"A digital sketch of the problem\" width=\"191\" height=\"300\" class=\"alignnone wp-image-2109 size-medium\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/helicopter-191x300.jpg 191w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/helicopter-65x102.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/helicopter.jpg 201w\" sizes=\"auto, (max-width: 191px) 100vw, 191px\" \/><\/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\/11\/helicopterFBD2.jpg\" alt=\"A free-body diagram of the problem\" width=\"201\" height=\"251\" class=\"alignnone wp-image-2112 size-full\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/helicopterFBD2.jpg 201w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/helicopterFBD2-65x81.jpg 65w\" sizes=\"auto, (max-width: 201px) 100vw, 201px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>m=5000 kg<\/li>\n<li><span>\u2220A = 40\u00b0<\/span><\/li>\n<li><span>\u2220B = 60\u00b0<\/span><\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>F<sub>1<\/sub><\/li>\n<li>F<sub>2\u00a0<\/sub><\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Considering the load is evenly distributed, the upper pair of string have equal lengths, and so does the lower pair, FBD can be drawn for each points. Use equilibrium equations for x and y direction to find tension on each string.<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>Weight,[latex]w =m\\times g= 5000 kg\\times 9.81= 49050 N[\/latex]<\/p>\n<p><span style=\"text-decoration: underline\">Point A<\/span><\/p>\n<p>[latex]\\begin{equation} \\sum{F_{y} = F_{1} \\sin{40} -\\dfrac{w}{4}} \\end {equation}[\/latex]<\/p>\n<p>[latex]\\begin{equation} F_{1} = \\dfrac{w}{4\\times \\sin{40}} = \\dfrac{49050}{4\\times 0.643} \\end{equation}[\/latex]<\/p>\n<p>[latex]F_{1} = 19070.76 N[\/latex]<\/p>\n<p><span style=\"text-decoration: underline\">Point B<\/span><\/p>\n<p>[latex]\\begin{equation} \\sum{F_{y} = F_{2} \\sin{60} -\\dfrac{w}{4}} \\end {equation}[\/latex]<\/p>\n<p>[latex]\\begin{equation} F_{2} = \\dfrac{w}{4\\times \\sin{60}} \\end{equation}[\/latex]<\/p>\n<p>[latex]F_{2} = 14159.51 N[\/latex]<\/p>\n<p><span style=\"text-decoration: underline\">Point C<\/span><\/p>\n<p>[latex]F_{T} =F_{1}\\sin{40} +F_{2}\\sin{60}[\/latex]<\/p>\n<p>[latex]\\begin{equation} \\quad\\quad\u00a0 = (19070.76\\times 0.643) +(14159.51\\times 0.866)\\end{equation}[\/latex]<\/p>\n<p>[latex]F_{T} = 24524.63 N[\/latex]<\/p>\n<p>Thus both the upper strings are at a tension of 19070.76 N and lower strings are at a tension of 14159.51 N.<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>Note that in this problem the vehicle&#8217;s weight was evenly distributed which is convenient in symmetrical situations like this. The tension on the strings can be verified as follows.<\/p>\n<p>[latex]2F_{T} = m\\times g[\/latex]<\/p>\n<\/div>\n<h1>Example 2.4.4: Equilibrium Equation, Submitted by Analiya Benny<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>A child with a mass of 20kg is sitting on a swing with both strings 4 m long.<\/p>\n<p>a. Draw the free-body diagram.<\/p>\n<p>b. Calculate the tension of the strings.<\/p>\n<p>c. What happens to the tension of the strings when string A is 3 m and string B is 4 m? Explain with a free-body diagram.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/child-girl-doll-happy-preview.jpg\" alt=\"Girl on a swing.\" width=\"499\" height=\"749\" class=\"alignnone wp-image-2059\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/child-girl-doll-happy-preview.jpg 728w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/child-girl-doll-happy-preview-200x300.jpg 200w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/child-girl-doll-happy-preview-683x1024.jpg 683w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/child-girl-doll-happy-preview-65x98.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/child-girl-doll-happy-preview-225x338.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/child-girl-doll-happy-preview-350x525.jpg 350w\" sizes=\"auto, (max-width: 499px) 100vw, 499px\" \/><\/p>\n<p>Image 1:\u00a0 A real-life demonstration of the situation in the question.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p><strong>2. Draw\u00a0<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-33-e1675991020826.png\" alt=\"A free-body diagram of the problem\" width=\"908\" height=\"468\" class=\"alignnone wp-image-2063 size-full\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-33-e1675991020826.png 908w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-33-e1675991020826-300x155.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-33-e1675991020826-768x396.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-33-e1675991020826-65x34.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-33-e1675991020826-225x116.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-33-e1675991020826-350x180.png 350w\" sizes=\"auto, (max-width: 908px) 100vw, 908px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>[latex]M= 20kg[\/latex]<\/li>\n<li>[latex]g = 9.81 m\/s^2[\/latex]<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>F<sub>A<\/sub><\/li>\n<li>F<sub>B<\/sub><\/li>\n<\/ul>\n<p><strong>4. Approach\u00a0<\/strong><\/p>\n<p>Analyzing the dimensions of the situation and applying the equilibrium equation.<\/p>\n<p><strong>5. Analysis\u00a0<\/strong><\/p>\n<p>[latex]w = m \\times g = 20 kg \\times 9.81= 196.2 N\u00a0[\/latex]<\/p>\n<p>b. When the length of the strings is equal, the weight on point A is equal to the weight on point B. F<sub>A<\/sub> = F<sub>B<\/sub><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-31-e1675989644427.png\" alt=\"A suspended cable system with endpoints A and B, 4 meters apart vertically, showing forces F_A and F_B each supporting half the total weight w.\" width=\"708\" height=\"420\" class=\"alignnone wp-image-2061 size-full\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-31-e1675989644427.png 708w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-31-e1675989644427-300x178.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-31-e1675989644427-65x39.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-31-e1675989644427-225x133.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-31-e1675989644427-350x208.png 350w\" sizes=\"auto, (max-width: 708px) 100vw, 708px\" \/><\/p>\n<p>No Force in the x-direction<\/p>\n<p>[latex]\\sum F_y = 0 \\\\  F_A + F_B = 196.2 \\, \\text{N} \\\\  F_B + F_B = 196.2 \\\\  2F_B = 196.2 \\\\  F_B = 98.1 \\, \\text{N} \\\\    \\text{Tension in each string is } 98.1 \\, \\text{N}[\/latex]<\/p>\n<p>c. When one chain of the swing is longer, the tension on the chains varies such that [latex]w_{B}>w_{A}[\/latex]; in turn, chain B will experience more tension than chain A. To solve for tension in the chains, we would require more information about the distribution of the weights on points A and B.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-32-e1675990212710.png\" alt=\"An asymmetric cable system with endpoints A and B at different heights, showing unequal weight distribution where w_B' &gt; w_A.\" width=\"816\" height=\"448\" class=\"alignnone wp-image-2062 size-full\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-32-e1675990212710.png 816w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-32-e1675990212710-300x165.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-32-e1675990212710-768x422.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-32-e1675990212710-65x36.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-32-e1675990212710-225x124.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/11\/Screenshot-32-e1675990212710-350x192.png 350w\" sizes=\"auto, (max-width: 816px) 100vw, 816px\" \/><\/p>\n<p><strong>6. Review\u00a0<\/strong><\/p>\n<p>In questions involving free-body diagrams, it is convenient to have perfectly symmetrical situations. However, when faced with non-symmetrical or distributed loads, different techniques need to be applied, which will be explained in the following chapters.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n","protected":false},"author":60,"menu_order":4,"template":"","meta":{"pb_show_title":"on","pb_short_title":"2.4 Examples","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-568","chapter","type-chapter","status-publish","hentry"],"part":52,"_links":{"self":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/568","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":27,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/568\/revisions"}],"predecessor-version":[{"id":2697,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/568\/revisions\/2697"}],"part":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/parts\/52"}],"metadata":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/568\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/media?parent=568"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapter-type?post=568"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/contributor?post=568"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/license?post=568"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}