{"id":569,"date":"2021-07-21T14:21:54","date_gmt":"2021-07-21T18:21:54","guid":{"rendered":"http:\/\/pressbooks.library.upei.ca\/statics\/?post_type=chapter&#038;p=569"},"modified":"2025-07-31T23:15:42","modified_gmt":"2025-08-01T03:15:42","slug":"3-6-examples","status":"publish","type":"chapter","link":"https:\/\/pressbooks.library.upei.ca\/statics\/chapter\/3-6-examples\/","title":{"raw":"3.6 Examples","rendered":"3.6 Examples"},"content":{"raw":"Here are examples from Chapter 3 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 3.6.1: Reaction Forces, Submitted by Andrew Williamson<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong><a href=\"https:\/\/www.google.com\/url?sa=i&amp;url=https%3A%2F%2Fwww.maxpixel.net%2FSeat-Couch-Interior-Home-Furniture-Room-Sofa-42817&amp;psig=AOvVaw1_RDGb7_NnDrMzVsitaA-r&amp;ust=1629764899959000&amp;source=images&amp;cd=vfe&amp;ved=0CAoQjRxqFwoTCICj1snxxfICFQAAAAAdAAAAABAD\" rel=\"https:\/\/www.google.com\/url?sa=i&amp;url=https%3A%2F%2Fwww.maxpixel.net%2FSeat-Couch-Interior-Home-Furniture-Room-Sofa-42817&amp;psig=AOvVaw1_RDGb7_NnDrMzVsitaA-r&amp;ust=1629764899959000&amp;source=images&amp;cd=vfe&amp;ved=0CAoQjRxqFwoTCICj1snxxfICFQAAAAAdAAAAABAD\"><\/a>\r\n<div class=\"textbox shaded\">\r\n\r\nA family is sitting watching TV on their couch. The couch is 5 m long and weighs 120 N. The child is sitting 1 m away from one end and has a mass of 30 kg. The mother is sitting 0.5 m away from the child and has a mass of 60 kg. The father is 3 m away from the mother and has a mass of 70 kg.\r\n\r\na) Draw a free-body diagram of the couch\r\n\r\nb) Calculate the reaction force on each of the two legs.\r\n\r\nAssume the couch is supported by two rollers.\r\n\r\n[caption id=\"attachment_1380\" align=\"aligncenter\" width=\"1024\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Seat-Couch-Interior-Home-Furniture-Room-Sofa-42817-1024x512.png\" alt=\"A couch\" class=\"wp-image-1380 size-large\" width=\"1024\" height=\"512\" \/> Source: https:\/\/www.maxpixel.net\/Seat-Couch-Interior-Home-Furniture-Room-Sofa-42817[\/caption]\r\n\r\n<\/div>\r\n\r\n<hr \/>\r\n\r\n<strong>2. Draw<\/strong>\r\n\r\nSketch:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Andrew-1-draw-1-300x262.jpg\" alt=\"A free body diagram showing three downward forces (Fc, FM, Ff) acting on a 5\u202fm beam, with distances marked as 1\u202fm, 0.5\u202fm, and 3\u202fm between them.\" class=\"aligncenter wp-image-844 size-medium\" width=\"300\" height=\"262\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>g = 9.81 m\/s<sup>2<\/sup><\/li>\r\n \t<li>m<sub>c<\/sub> = 30 kg<\/li>\r\n \t<li>m<sub>m<\/sub> = 60 kg<\/li>\r\n \t<li>m<sub>f<\/sub> = 70 kg<\/li>\r\n \t<li>F<sub>g<\/sub> = 120 N<\/li>\r\n \t<li>r<sub>c<\/sub> = 1 m<\/li>\r\n \t<li>r<sub>M<\/sub> = 1.5 m<\/li>\r\n \t<li>r<sub>f<\/sub> = 4.5 m<\/li>\r\n \t<li>r<sub>B<\/sub> = 5 m<\/li>\r\n \t<li>r<sub>g<\/sub> = 2.5 m<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>N<sub>A<\/sub><\/li>\r\n \t<li>N<sub>B<\/sub><\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nUse equilibrium equations\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\nPart a:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/andrew-1-draw-2-300x282.jpg\" alt=\"Free body diagram of a beam with applied forces (Fc, FM, Ff, Fg), reaction forces (NA, NB), and position vectors (rC, rM, rG, rF, rB) shown from point A.\" class=\"alignnone wp-image-937\" width=\"369\" height=\"347\" \/>\r\n\r\nPart b:\r\n\r\n[latex]\\sum M_A=0=N_B\\times r_B-F_c\\times r_c-F_M\\times r_M-F_f\\times r_f-F_g\\times r_g\\\\\\\\N_B(5m)=(30kg\\times 9.81m\/s^2)(1m)+(60kg\\times 9.81m\/s^2)(1.5m)\\\\+(70kg\\times 9.81m\/s^2)(4.5m)+(120N)(2.5m)[\/latex][latex]\\\\N_B(5m)=294.3Nm+882.9Nm +3090.15Nm+300Nm\\\\\\\\N_B(5m)=4567.35N m\\\\\\\\N_B\\frac{4567.35Nm}{5m}\\\\\\\\N_B=913.47N\\\\\\\\N_B=913N[\/latex]\r\n\r\n[latex]\\sum F_y=0=N_A+N_B-F_C-F_M-F_f-F_g\\\\\\\\N_A=F_C+F_M+F_f+F_g-N_B\\\\\\\\N_A=(30kg\\times 9.81m\/s^2)+(60kg\\times 9.81m\/s^2)\\\\+(70kg\\times 9.81m\/s^2)+120N-913.47N[\/latex] [latex]\\\\N_A=294.3N+588.6N+686.7N+120N-913.47N$$$$\\\\N_A=776.13N\\\\\\\\\\\\\\underline{N_A=776N}[\/latex]\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nIt is interesting that N<sub>B<\/sub> is larger than N<sub>A<\/sub>, because the weight of the mother and child combined (80 kg) is larger than that of the father (70 kg). However, when you sum the moments at point B instead of A, you get the same answer. The distance between the reaction forces and the nearest forces is important, as well as the magnitude of the forces themselves. The distance between A and F<sub>c<\/sub> is 1 m, while the distance between B and F<sub>f<\/sub> is only 0.5 m.\r\n\r\nAdditionally, it makes sense that both N<sub>A<\/sub> and N<sub>B<\/sub> are positive, i.e. are in the positive y direction.\r\n\r\n<\/div>\r\n<h1>Example 3.6.2: Couples, Submitted by Kirsty MacLellan<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nA water valve is opened by a wheel with a diameter of 10 inches. It takes 7.5 lb of force to open the valve. What is the moment it takes to open the valve?\r\n\r\nReal-life scenario:\r\n\r\n[caption id=\"attachment_1397\" align=\"aligncenter\" width=\"910\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/wheel-valve-heating-line.jpg\" alt=\"Three large gate valves\" class=\"wp-image-1397 size-full\" width=\"910\" height=\"640\" \/> Source: https:\/\/www.pxfuel.com\/en\/free-photo-ekahu[\/caption]\r\n\r\n<\/div>\r\n\r\n<hr \/>\r\n\r\n<strong>2. Draw<\/strong>\r\n\r\nSketch:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Kirsty-1-draw-1-300x281.jpg\" alt=\"10-inch wheel with opposite forces causing rotation.\" class=\"alignleft wp-image-926\" width=\"315\" height=\"295\" \/>\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\nFree-body diagram:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Kirsty-1-draw-2-1-300x224.jpg\" alt=\"A FBD of a wheel with opposing vertical forces F_A and F_B, and moment arms r_A, r_B.\" class=\"alignnone wp-image-934\" width=\"295\" height=\"220\" \/>\r\n\r\n&nbsp;\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnown:\r\n<ul>\r\n \t<li>d = 10 in<\/li>\r\n \t<li>F = 7.5 lb<\/li>\r\n<\/ul>\r\nUnknown:\r\n<ul>\r\n \t<li><span style=\"text-decoration: underline\">M<\/span><\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nDetermine the moment by finding the cross product of <span style=\"text-decoration: underline\">r<\/span><sub>A<\/sub> and <span style=\"text-decoration: underline\">F<\/span><sub>A<\/sub>, then <span style=\"text-decoration: underline\">r<\/span><sub>B<\/sub> and <span style=\"text-decoration: underline\">F<\/span><sub>B<\/sub>, then adding.\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\nFind radius:\r\n\r\n[latex]r=\\frac{d}{2}\\\\r=\\frac{10in}{2}\\\\r=5in\\\\5in\\times\\frac{1ft}{12in}=0.42ft[\/latex]\r\n\r\nFind <span style=\"text-decoration: underline\">r<\/span><sub>A<\/sub>, <span style=\"text-decoration: underline\">F<\/span><sub>A<\/sub>, <span style=\"text-decoration: underline\">r<\/span><sub>B<\/sub>, and <span style=\"text-decoration: underline\">F<\/span><sub>B<\/sub> in vector form:\r\n\r\n[latex]\\underline{r}_A= \\begin{bmatrix}\r\n0.42 \\\\\r\n0\r\n\\end{bmatrix}ft\\:\\; \\underline{F}_A=\\begin{bmatrix}\r\n0 \\\\\r\n7.5\r\n\\end{bmatrix}lb \\\\\\underline{r}_B=\\begin{bmatrix}\r\n-0.42 \\\\\r\n0\r\n\\end{bmatrix}ft\\:\\; \\underline{F}_B=\\begin{bmatrix}\r\n0 \\\\\r\n-7.5\r\n\\end{bmatrix}lb[\/latex]\r\n\r\nFind <span style=\"text-decoration: underline\">M<\/span><sub>A<\/sub>:\r\n\r\n[latex]\\underline{M}_A=\\underline{r}_A\\times \\underline{F}_A=\\begin{bmatrix}\r\n\\underline{\\hat{i}} &amp; \\underline{\\hat{j}} &amp; \\underline{\\hat{k}} \\\\\r\n0.42 &amp; 0 &amp; 0 \\\\\r\n0 &amp; 7.5 &amp; 0\r\n\\end{bmatrix}[\/latex] [latex]\\underline{M}_A=\\hat{i} \\begin{bmatrix}\r\n0 &amp; 0 \\\\\r\n7.5 &amp; 0\r\n\\end{bmatrix} -\\underline{\\hat{j}} \\begin{bmatrix}\r\n0.42 &amp; 0 \\\\\r\n0 &amp; 0\r\n\\end{bmatrix}+\\underline{\\hat{k}} \\begin{bmatrix}\r\n0.42 &amp; 0 \\\\\r\n0 &amp; 7.5\r\n\\end{bmatrix}\\\\\\underline{M}_A=(\\underline{\\hat{i}}(0)-\\underline{\\hat{j}}(0)+\\underline{\\hat{k}}(0.42\\cdot 7.5-0\\cdot 0))ft\\cdot lb\\\\\\underline{M}_A=3.15\\underline{\\hat{k}} ft\\cdot lb[\/latex]\r\n\r\nFind <span style=\"text-decoration: underline\">M<\/span><sub>B<\/sub>:\r\n\r\n[latex]\\underline{M}_B=\\underline{r}_B\\times \\underline{F}_B=\\begin{bmatrix}\r\n\\underline{\\hat{i}} &amp; \\underline{\\hat{j}} &amp; \\underline{\\hat{k}} \\\\\r\n-0.42 &amp; 0 &amp; 0 \\\\\r\n0 &amp; -7.5 &amp; 0\r\n\\end{bmatrix}[\/latex] [latex]\\underline{M}_B=\\hat{i} \\begin{bmatrix}\r\n0 &amp; 0 \\\\\r\n-7.5 &amp; 0\r\n\\end{bmatrix} -\\underline{\\hat{j}} \\begin{bmatrix}\r\n-0.42 &amp; 0 \\\\\r\n0 &amp; 0\r\n\\end{bmatrix}+\\underline{\\hat{k}} \\begin{bmatrix}\r\n-0.42 &amp; 0 \\\\\r\n0 &amp; -7.5\r\n\\end{bmatrix}\\\\\\underline{M}_B=(\\underline{\\hat{i}}(0)-\\hat{j}(0)+\\underline{\\hat{k}}(-0.42\\cdot -7.5-0\\cdot 0))ft\\cdot lb\\\\\\underline{M}_B=3.15\\underline{\\hat{k}} ft\\cdot lb[\/latex]\r\n\r\nAdd <span style=\"text-decoration: underline\">M<\/span><sub>A<\/sub> and <span style=\"text-decoration: underline\">M<\/span><sub>B<\/sub> to get <span style=\"text-decoration: underline\">M<\/span>:\r\n\r\n[latex]\\underline{M}=\\underline{M}_A+\\underline{M}_B\\\\\\underline{M}=3.15ft\\cdot lb+3.15ft\\cdot lb\\\\\\underline{M}=6.3\\underline{\\hat{k}}ft\\cdot lb[\/latex]\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nNotice that for the x-coordinates, the positive x-direction was taken to the left, and this was consistently followed in the calculation. It's important to stay consistent with your chosen direction, as mixing it up can lead to sign errors.\r\n\r\nThis answer makes sense because there is only one moment acting in the k direction.\r\n\r\nNote: We could have come to the same answer using the formula M = F*d, which would have been faster.\r\n\r\n[latex]M=f\\cdot d\\\\M=7.5lb\\cdot 10in(\\frac{1ft}{12in})\\\\M=7.5lb\\cdot \\frac{5}{6}ft\\\\\\\\M=6.25ft\\cdot lb[\/latex]\r\n\r\nThis answer is slightly more accurate because we didn't round when converting between inches and feet (in the original solution, we rounded 0.416667 to 0.42).\r\n\r\n<\/div>\r\n<h1>Example 3.6.3: Distributed Load, Submitted by Luciana Davila<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nA shelf on the wall is 1.5 meters away from the floor. The shelf has a length of 100 cm. A person starts putting different objects on it to create a distributed load. The load created a curve described by:\r\n<p style=\"text-align: center\">w = 4x<sup>4<\/sup> +2 N\/m.<\/p>\r\nCalculate the resultant force and how far it is acting from the wall (Fixed end).\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"300\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/shelf-576088_1280-300x150.png\" alt=\"A shelf\" width=\"300\" height=\"150\" style=\"border: 1px solid black\" \/> Source: https:\/\/pixabay.com\/vectors\/shelf-floating-bathroom-glass-576088\/[\/caption]\r\n\r\n<\/div>\r\n\r\n<hr \/>\r\n\r\n<strong>2. Draw<\/strong>\r\n\r\nSketch:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luciana-1-draw-2-275x300.jpg\" alt=\"A triangular distributed load on a cantilever beam, 100\u202fcm long and 1.5\u202fm high.\" class=\"alignleft wp-image-945 size-medium\" width=\"275\" height=\"300\" \/>\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\nFree-body diagram:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luciana-1-draw-1-2-300x156.jpg\" alt=\"Free body diagram of a cantilever beam with a variable load w = 4x^4 + 2, showing reaction forces and moment at the fixed end.\" class=\"alignnone wp-image-947\" width=\"440\" height=\"229\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>w = 4x<sup>4<\/sup> + 2 N\/m<\/li>\r\n \t<li>L = 100 cm = 1m<\/li>\r\n \t<li>xmin = 0m<\/li>\r\n \t<li>xmax = 1m<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>x<sub>r<\/sub><\/li>\r\n \t<li>F<sub>r<\/sub><\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nUse distributed load equations:\r\n\r\n[latex]F_r=\\int^{xmax}_{xmin} wdx\\\\X_r=\\frac{\\int^{xmax}_{xmin} x*w(x)*dx}{\\int^{xmax}_{xmin}wdx}[\/latex]\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\nSolve for F<sub>r<\/sub>:\r\n\r\n[latex]F_r=\\int^1_0 (4x^4+2)dx\\;\\; N\\\\F_r=(\\frac{4x^5}{5}+2x)\\vert^1_0\\;\\;N\\\\F_r=(\\frac{4}{5}+2)N\\\\F_r=2.8N[\/latex]\r\n\r\nSolve for x<sub>r<\/sub>:\r\n\r\n[latex]X_r=\\frac{(\\int^1_0x(4x^4+2)dx)N\/m}{\\int^1_0(4x^4+2)dx)N}\\\\X_r=\\frac{\\int^1_0(4x^5+2x)dxN\/m}{2.8N}\\\\X_r=\\frac{(\\frac{4x^6}{6}+\\frac{2x^2}{2})\\vert^1_0N\/m}{2.8N}\\\\X_r=\\frac{(\\frac{2}{3}+1)N\/m}{2.8N}\\\\X_r=0.59m[\/latex]\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nThe function shows an increasing curve on the interval, so it makes sense that the resultant force would be applied closer to the right end of the beam than the left end.\r\n\r\n<\/div>\r\n<h1>Example 3.6.4: Couple, Submitted by Hamza <span>Ben Driouech<\/span><\/h1>\r\n<div class=\"textbox\">\r\n<ol>\r\n \t<li><strong>Problem<\/strong><\/li>\r\n<\/ol>\r\n<div class=\"textbox shaded\">\r\n\r\nA mason jar lid with a diameter of 11cm is firmly closed and requires a moment of 5.5 Nm to be opened. If you are to open the lid with your finger and thumb, assuming they are applying equal force, determine the force required by your thumb and finger.\r\n\r\n[caption id=\"attachment_1847\" align=\"aligncenter\" width=\"409\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/jar-lid-opening.jpg\" alt=\"A person tightly gripping and twisting the lid of a jar to open it.\" class=\"wp-image-1847\" width=\"409\" height=\"318\" \/> Source: https:\/\/commons.wikimedia.org\/wiki\/File:Fingertip_tightening_lid_on_pickled_beet_jar.jpg[\/caption]\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\/2022\/02\/Screenshot-2025-05-07-140431-223x300.png\" alt=\"A free-body diagram of a jar lid showing a 5.5 Nm moment.\" width=\"223\" height=\"300\" style=\"border: 1px solid black\" class=\"alignnone\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>M = 5.5 Nm<\/li>\r\n \t<li>d = 11 cm = 0.11 m<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>F - Force exerted by thumb and finger<\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nUsing couple to find F\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\nSince the forces exerted by the finger and thumb are equal and opposite, it is possible to find the force exerted by the finger and thumb using the couple equation.\r\n\r\n[latex]5.5Nm=F\\times0.11m[\/latex]\r\n\r\n[latex]F=\\frac{5.5Nm}{0.11m}\\\\F =50N[\/latex]\r\n\r\nThus, your thumb and finger are applying a force of 50N each.\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nThe moment equation is applicable since the force of the thumb and finger is equal. The equation produces a reasonable value of 50N\r\n\r\n<\/div>\r\n<h1>Example 3.6.5: Couple, Submitted by Andrew Williamson.<\/h1>\r\n<div class=\"textbox\">\r\n<ol>\r\n \t<li><strong>Problem<\/strong><\/li>\r\n<\/ol>\r\n<div class=\"textbox shaded\">\r\n\r\nYou are driving home from work, and you need to turn right to get into your driveway. To turn the steering, you are applying 115 N with each hand. Consider your steering wheel has a diameter of 40 cm.\r\n\r\na. Determine the couple moment produced on the steering wheel.\r\n\r\nb. Compare the above result with a situation where you use single hand to navigate the wheel. How much force does it take to turn the steering wheel in this case?\r\n\r\n[caption id=\"attachment_1836\" align=\"alignnone\" width=\"404\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Andrew-williamson-wheel-problem.jpg\" alt=\"A person turning a steering wheel\" class=\"wp-image-1836\" width=\"404\" height=\"269\" \/> Source:https:\/\/commons.wikimedia.org\/wiki\/File:Woman_hand_steering_wheel_%28Unsplash%29.jpg[\/caption]\r\n\r\n&nbsp;\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/hands-on-wheel-andrew-williamson-e1652710940105.jpg\" alt=\"A sketch of a person turning the steering wheel.\" class=\"alignnone wp-image-1837\" width=\"326\" height=\"271\" \/>\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\/andrew-williamson-steeering-wheel-6.jpg\" alt=\"Two steering wheel setups: one with opposite 115\u202fN forces creating a moment, the other with one angled force F_2.\" class=\"alignnone wp-image-1846 size-full\" width=\"701\" height=\"334\" \/>\r\n\r\n<strong><span style=\"text-align: initial;font-size: 1em\">3. Knowns and Unknowns<\/span><\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>F = 115 N<\/li>\r\n \t<li>d = 40 cm = 0.40m<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>Couple moment, M<\/li>\r\n \t<li>Force applied by hand in the second scenario, F<sub>2<\/sub><\/li>\r\n<\/ul>\r\n<strong>4. Approach<\/strong>\r\n\r\nUse the couple Moment equation to find the moment, and then use the moment to find the force required to turn the handle single-handedly.\r\n\r\n<strong>5. Analysis<\/strong>\r\n\r\na.\u00a0 Both arms are applying a 115 N force each. The diameter of the wheel is 40 cm.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/hands-on-wheel-andrew-williamson-e1652710940105.jpg\" alt=\"A sketch of a person turning the steering wheel.\" class=\"alignnone wp-image-1837 size-full\" width=\"225\" height=\"187\" \/><img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/andrew-williamson-steeering-wheel-2.jpg\" alt=\"Two opposite 115\u202fN forces act on a 0.4\u202fm diameter wheel to create a moment.\" class=\"alignnone wp-image-1840 size-full\" width=\"306\" height=\"293\" \/>\r\n\r\nTherefore,\r\n\r\n[latex]\\overrightarrow{M}= \\overrightarrow{F}\\times\\overrightarrow{d}=115N\\times0.4m[\/latex]\r\n\r\n[latex]\\overrightarrow{M}=46Nm[\/latex]\r\n\r\nb. Consider 46Nm as the required moment to turn the steering wheel.\r\n\r\nWhen just one hand applies force on the wheel it forms a moment which is equal to 46 Nm, with distance and angle given we can find the force it takes to turn the steering wheel.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/andrew-williamson-steeering-wheel-4.jpg\" alt=\"Force F_2 is applied to a 0.4\u202fm diameter wheel to produce a moment.\" class=\"aligncenter wp-image-1842\" width=\"288\" height=\"249\" \/>\r\n\r\n[latex]\\overrightarrow{M} = F_{2}\\times\\frac{d}{2}\\times\\sin{\\theta}\\\\46Nm =F_{2}\\cdot\\frac{0.4m}{2}\\cdot\\sin{90}\\\\F_{2}=230 N[\/latex]\r\n\r\n<strong>6. Review\u00a0<\/strong>\r\n\r\nAs expected, the force it took to turn the steering wheel is twice the force it took with one hand when both hands are used.\r\n\r\n<\/div>\r\n<h1>Example 3.6.6: Couple, Submitted by Elliot Fraser<\/h1>\r\n<div class=\"textbox\">\r\n<ol>\r\n \t<li><strong>Problem<\/strong><\/li>\r\n<\/ol>\r\n<div class=\"textbox shaded\">\r\n\r\nEllen is taking her driving test and is asked to perform a right turn. The diameter of the steering wheel is 15 inches. Ellen's right hand exerts a force of 200 N, and her left exerts 150 N.\r\n\r\na. Calculate the net moment in the above-given situation using the cross product.\r\n\r\nb. If Ellen were to perform a left turn with the same forces given, what would be the net moment?\r\n\r\n[caption id=\"attachment_1901\" align=\"alignnone\" width=\"705\"]<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Elliot-fraser-steering-wheel.jpg\" alt=\"A person turning a Toyota steering wheel with both hands.\" class=\"wp-image-1901\" width=\"705\" height=\"470\" \/> Source:https:\/\/www.wikihow.com\/Drive-Safely-in-Fog[\/caption]\r\n\r\nThe above image is a close illustration of the problem. In the problem, the hands of the driver are along the diameter.\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\/2022\/02\/Elliot-fraser-chap-3-1.jpg\" alt=\"Diagram showing right and left steering turns with opposing forces applied 15 inches apart.\" class=\"alignnone wp-image-1905 size-full\" width=\"726\" height=\"426\" \/>\r\n\r\n&nbsp;\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>Diameter of the wheel = 15 in = 0.381m<\/li>\r\n \t<li>Right hand force = 200 N<\/li>\r\n \t<li>left hand force = 150 N<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>M<sub>1\u00a0<\/sub>(for right hand)<\/li>\r\n \t<li>M<sub>2 <\/sub>(for left hand)<\/li>\r\n<\/ul>\r\n<strong>4. Approach\u00a0<\/strong>\r\n\r\nConvert inches to meters and add the cross product of distance and force to find the net moment.\r\n\r\n<strong>5. Analysis\u00a0<\/strong>\r\n\r\na.RIGHT TURN\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-21.png\" alt=\"A diagram of a right turn showing a couple: 150\u202fN upward and 200\u202fN downward, 15\u202fin apart.\" class=\"alignnone wp-image-1912 size-full\" width=\"383\" height=\"414\" \/>\r\n\r\nExpress the radius of the steering wheel and forces in vector form for tight turn.\r\n\r\n[latex]\\overrightarrow {r_{1}} = \\begin{bmatrix}-0.1905\\\\0\\\\0\\end{bmatrix}\\kern 1pc\\overrightarrow {r_{2}} = \\begin{bmatrix}0.1905\\\\0\\\\0\\end{bmatrix}[\/latex]\r\n\r\n[latex]\\overrightarrow {F_{1}} = \\begin{bmatrix}0\\\\150\\\\0\\end{bmatrix}\\kern 1pc\\overrightarrow {F_{2}} = \\begin{bmatrix}0\\\\-200\\\\0\\end{bmatrix}[\/latex]\r\n\r\n[latex]\\overrightarrow{M_1}=\\overrightarrow{r_{1}} \\times \\overrightarrow{F_{1}} \\kern 1pc +\u00a0 \\kern 1pc\\overrightarrow{r_{2}} \\times \\overrightarrow{F_{2}}[\/latex]\r\n\r\n<span style=\"text-align: initial;font-size: 1em\">[latex]\\overrightarrow{M_{1}}=\\:\\begin{vmatrix} \\hat{i} &amp; \\hat{j} &amp; \\hat{k} \\\\-0.1905 &amp; 0 &amp; 0\\\\0 &amp;\u00a0 150 &amp; 0\\end{vmatrix}[\/latex][latex]\\kern 1pc + \\kern 1pc\u00a0 \\begin{vmatrix} \\hat{i} &amp; \\hat{j} &amp; \\hat{k} \\\\0.1905 &amp; 0 &amp; 0\\\\0 &amp;\u00a0 -200\u00a0 &amp; 0\\end{vmatrix}[\/latex][latex]\\:=-28.575\\: \\hat{k} - 38.1 \\hat{k}[\/latex]<\/span>\r\n\r\n[latex]\\overrightarrow{M_{1}}=\\: -66.675 \\: \\hat{k}[\/latex]\r\n\r\nb. LEFT TURN\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-22.png\" alt=\"A diagram of a left turn showing a couple: 200\u202fN upward and 150\u202fN downward, 15\u202fin apart.\" class=\"alignnone wp-image-1913 size-full\" width=\"361\" height=\"437\" \/>\r\n\r\nRadius and force in vector form\r\n\r\nRadius is the same as it is for a right turn, but the forces have a different direction.\r\n\r\n[latex]\\overrightarrow {r_{1}} = \\begin{bmatrix}-0.1905\\\\0\\\\0\\end{bmatrix}\\kern 1pc\\overrightarrow {r_{2}} = \\begin{bmatrix}0.1905\\\\0\\\\0\\end{bmatrix}[\/latex]\r\n\r\n[latex]\\overrightarrow {F_{3}} = \\begin{bmatrix}0\\\\-150\\\\0\\end{bmatrix}\\kern 1pc\\overrightarrow {F_{4}} = \\begin{bmatrix}0\\\\200\\\\0\\end{bmatrix}[\/latex]\r\n\r\n[latex]\\overrightarrow{M_2}=\\overrightarrow{r_{1}} \\times \\overrightarrow{F_{3}} \\kern 1pc +\u00a0 \\kern 1pc\\overrightarrow{r_{2}} \\times \\overrightarrow{F_{4}}[\/latex]\r\n\r\n<span style=\"text-align: initial;font-size: 1em\">[latex]\\overrightarrow{M_{2}}=\\:\\begin{vmatrix} \\hat{i} &amp; \\hat{j} &amp; \\hat{k} \\\\-0.1905 &amp; 0 &amp; 0\\\\0 &amp;\u00a0 -150 &amp; 0\\end{vmatrix}[\/latex][latex]\\kern 1pc + \\kern 1pc\u00a0 \\begin{vmatrix} \\hat{i} &amp; \\hat{j} &amp; \\hat{k} \\\\0.1905 &amp; 0 &amp; 0\\\\0 &amp;\u00a0 200\u00a0 &amp; 0\\end{vmatrix}[\/latex][latex]\\:=28.575\\: \\hat{k} + 38.1 \\hat{k}[\/latex]<\/span>\r\n\r\n&nbsp;\r\n\r\n[latex]\\overrightarrow{M_{2}}=\\: 66.675 \\: \\hat{k}[\/latex]\r\n\r\n<strong>6. Review\u00a0<\/strong>\r\n\r\nBecause the radius and force values in both instances have the same magnitude but different directions, the net moments for left and right turns were expected to have the same magnitudes and different directions; thus, the answers make sense.\r\n\r\n<\/div>\r\n<h1 id=\"chapter-1676-section-1\" class=\"section-header\">Example 3.6.7: Lever Arms, Submitted by Dhruvil Kanani<\/h1>\r\n<div class=\"textbox\">\r\n\r\n1. <strong>Problem<\/strong>\r\n<div class=\"textbox shaded\">Josh has two sons; the older one weighs 50 kg, while the younger one weighs 35 kg. Josh wants to\r\nbuild a Seesaw (tatter-totter) for his sons in a way that both the kids have to apply an equal amount\r\nof force. What is the required ratio of the lengths of both arms to achieve this? Acceleration due to gravity is 9.81m\/s<sup>2<\/sup>.<\/div>\r\n<strong>2. Sketch:<\/strong>\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Chapter3_Dhru_1-1-300x136.png\" alt=\"A simply supported beam with loads applied at different positions along its length.\" width=\"300\" height=\"136\" class=\"alignnone wp-image-2120 size-medium\" \/><img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Chapter3_Dhru_2-1-300x125.png\" alt=\"A free body diagram of the problem.\" width=\"300\" height=\"125\" class=\"alignnone wp-image-2121 size-medium\" \/>\r\n<strong>3. Knowns and Unknowns:<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li><span>m<\/span><sub>1<\/sub><span>\u00a0= 35kg<\/span><\/li>\r\n \t<li><span>m<\/span><sub>2<\/sub><span> = 50kg<\/span><\/li>\r\n \t<li><span>g = 9.81 m\/s<\/span><sup>2<\/sup><\/li>\r\n<\/ul>\r\nUnknown:\r\n<ul>\r\n \t<li><span>R<\/span><sub>1<\/sub><\/li>\r\n \t<li><span>R<\/span><sub>2<\/sub><\/li>\r\n<\/ul>\r\n<strong style=\"text-align: initial;background-color: initial;font-size: 1em\">4. Approach:<\/strong>\r\n\r\nConvert the masses to forces using gravity<strong>.\r\n<\/strong>Equate the moments caused by each force.\r\nSolve for the ratio of lengths.\r\n\r\n<strong>5. Analysis:<\/strong>\r\nFor the children to apply an equal amount of force while playing on the seesaw, they must create an equal moment.\r\nTherefore:\r\n[latex]M_1 = M_2[\/latex]\r\nThese moments can be rewritten in terms of F and R.\r\n[latex]F_1 \\cdot R_1 = F_2 \\cdot R_2[\/latex]\r\nThe F values can be solved by multiplying the child's mass by gravity to find the force they exert on the seesaw.\r\n[latex] (m_1 \\cdot g) \\cdot R_1 = (m_2 \\cdot g) \\cdot R_2[\/latex]\r\nGravity can be cancelled out as it exists on both sides of the equation, leaving:\r\n[latex]m_1\u00a0 \\cdot R_1 = m_2 \\cdot R_2[\/latex]\r\nPutting values into this equation gives:\r\n[latex]35kg\u00a0 \\cdot R_1 = 50kg \\cdot R_2[\/latex]\r\nThe equation can now be adjusted to solve for R<sub>1<\/sub>\/R<sub>2<\/sub>.\r\n[latex]\\frac{R_1}{R_2} = \\frac{50kg}{35kg}[\/latex]\r\nThis can be simplified to give the ratio:\r\n[latex]\\frac{R_1}{R_2} = \\frac{10}{7} \\\\ R_1 : R_2 = 10 : 7[\/latex]\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nIt makes sense that the lighter child needs to be further away to produce an equal moment to that of the heavier child. 10:7 is a reasonable ratio that matches the ratio of their weights.\r\n\r\n<\/div>\r\n<h1 id=\"chapter-1676-section-1\" class=\"section-header\">Example 3.6.8: Distributed loads, Submitted by Will Craine<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nThere are 5 cars on a bridge, three in one lane and the rest in the second lane. Each car is of identical dimensions and weight.\r\n\r\na. Find the resultant force of the cars from the side of the bridge with 3 cars.\r\n\r\nb. Find the resultant force of the cars from the side of the bridge, with 2 cars in the opposite direction to the other 3 cars.\r\n\r\nc. Find the resultant force of all 5 cars and the location of this force using the average of the x and y coordinates.\r\n<p style=\"text-align: center\"><img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Updates-car-300x163.jpg\" alt=\"Layout of objects A to E in a 13\u202fm wide area.\" width=\"491\" height=\"267\" class=\"aligncenter wp-image-2125\" \/><span style=\"text-align: initial;font-size: 1em\">Figure 1: Top view of the bridge\u00a0<\/span><\/p>\r\n&nbsp;\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-20-1-e1674753024654.png\" alt=\"A 3D load layout over a 13\u202fm span, and 1000\u202fN\/m triangular distributed loads.\" width=\"574\" height=\"386\" class=\"aligncenter wp-image-2033\" \/>\r\n<p style=\"text-align: center\">Figure 2: Drawing denoting direction and dimension of cars and bridge (Assuming the cars are in a Trapezium Shape).<\/p>\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\/2022\/02\/Screenshot-20-1-e1674753024654.png\" alt=\"A 3D load layout over a 13\u202fm span, and 1000\u202fN\/m triangular distributed loads.\" width=\"479\" height=\"322\" class=\"alignnone wp-image-2033\" \/>\r\n\r\n<strong>3. Knowns and Unknowns<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>Dimensions of the cars and the bridge<\/li>\r\n \t<li>Weight of one car, w = 1000 N\/m<\/li>\r\n<\/ul>\r\nUnknowns:\r\n<ul>\r\n \t<li>The resultant force on either side of the bridge.<\/li>\r\n \t<li>The location of the resultant force of all 5 cars.<\/li>\r\n<\/ul>\r\n<strong>4.\u00a0 Approach<\/strong>\r\n<ul>\r\n \t<li>The weight of the cars is distributed along the dimensions. We figure out he resultant forces and the location of the resultant forces along the x direction using the equation [latex]\\bar{x}=\\frac{\\sum F_{i}x_i}{\\sum F_i}[\/latex].<\/li>\r\n \t<li>Note that the shape of the car consists of a rectangle and an adjacent triangle. Hence, the distributed load is found for both shapes, using equations appropriate to find the area under the force, and then added to form the resultant force of one car.<\/li>\r\n<\/ul>\r\n<strong>5. Analysis<\/strong>\r\n\r\na. <span style=\"text-decoration: underline\">Finding the location along the x-axis and the magnitude of the resultant force of 3 cars<\/span>\r\n\r\nSince all 3 cars are identical [latex]F_{A}=F_{B}=F_{C}[\/latex].\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-21-1-e1674754059542.png\" alt=\"Triangular uniform distributed loads over a 3\u202fm span with 1000\u202fN\/m.\" width=\"604\" height=\"232\" class=\"alignnone wp-image-2036 size-full\" \/>\r\n\r\nFrom the fig above\r\n\r\n[latex]F_{R1}=1000 N\/m \\times 1.5 m = 1500 N [\/latex]\r\n\r\n[latex]F_{R2} = \\frac{1}{2} \\times 1.5 m \\times 1000 N\/m = 750 N[\/latex]\r\n\r\n[latex] F_{A} = 1500 N + 750 N = 2250N[\/latex]\r\n\r\n<span style=\"text-decoration: underline\">[latex]\\bar {x} [\/latex] for resultant force of car A<\/span>\r\n\r\n&nbsp;\r\n\r\nFrom O [latex]F_{R1}[\/latex] locates at [latex]1+ \\frac{1}{2}(1.5) = 1.75 m [\/latex]\r\n\r\n[latex]F_{R2}[\/latex] locates at [latex]1+ 1.5+ \\frac{1}{3}(1.5) = 3 m [\/latex]\r\n\r\nnow, [latex]\\bar{x}=\\frac{\\sum F_{i}x_i}{\\sum F_i}[\/latex]\r\n\r\nthus [latex]\\bar{x_{A}}= \\frac {(1500 N\/m\\times 1.75m+(750 N\/m\\times 3m)}{(1500N +750N)} [\/latex]\r\n\r\n[latex]\\bar{x_{A}} = 2.17 m [\/latex] from left side.\r\n\r\nAlso for one car the [latex]\\bar{x}[\/latex] of 2250 N force is [latex]2.17m - 1 m = 1.17 m [\/latex]\r\n\r\nUsing these calculations, the following figure can be drawn.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-24-e1674755299302.png\" alt=\"Locations of forces A, B, and C with their x-coordinates.\" width=\"800\" height=\"216\" class=\"alignnone wp-image-2039 size-full\" \/>\r\n\r\nThe total resultant force of all 3 cars is [latex]F_{R3} = 2250N \\times 3 = 6750 N [\/latex]\r\n\r\n[latex]\\bar{x}=\\frac{\\sum F_{i}x_i}{\\sum F_i}[\/latex]\r\n\r\ntherefore,\r\n\r\n[latex]\\bar{x} = \\frac{(2.17 \\times 2250)+(6.17\\times 2250)+(10.17\\times 2250)}{6750}[\/latex]\r\n\r\n[latex]\\bar{x} = 6.17 m [\/latex]\u00a0 from left side .\r\n\r\n&nbsp;\r\n\r\nb. <span style=\"text-decoration: underline\">Position and magnitude of the resultant force of the 2 cars on the lane.\u00a0<\/span>\r\n\r\n[latex]F_{D} = F_{E}=2250 N[\/latex]\r\n\r\nTotal force of the 2 cars [latex]F_{R2} = 2250 \\times 2= 4500 N [\/latex]\r\n\r\nPosition of the resultant forces from the 2 cars,\r\n\r\n[latex]\\bar {x_{2}} = \\frac {(4.17 \\times 2250) + (9.17 \\times 2250)}{4500}=6.67 m [\/latex]\r\n\r\n6.67 m is from the right side, but from the\u00a0 left side\u00a0 [latex]\\bar{x_{2}} = 13-6.67 = 6.33 m [\/latex]\r\n\r\nc. Position and magnitude of the total resultant force on the bridge.\r\n\r\nTotal force on bridge due to the 5 cars [latex]F_{R5} =F_{R2} +F_{R3} = 4500 N + 6750 N = 11250 N [\/latex]\r\n\r\n[latex]\\bar{x} = \\frac{(F_{R2}\\times \\bar{x_{2}})+(F_{R3}\\times\\bar{x_{3}})}{F_{R5}}[\/latex]\r\n\r\n[latex]\\bar {x} =\\frac{ (6.33 \\times 4500)+ (6.17 \\times 6750 )}{11250}= 6.234 m[\/latex]\r\n\r\n[latex]\\bar {y}\u00a0 [\/latex] for all the cars is half of its width\r\n\r\ntherefore, [latex]\\bar{y_{5}} = \\frac{(\\bar{y_{2}}\\times F_{R3} )+(\\bar{}y_{3}\\ times F_{R3})}{F_{R5}}[\/latex]\r\n\r\n[latex]\\bar{y_{5}} = \\frac{(6m \\times 4500 N )+(2m \\times 6750 N)}{11250 N}[\/latex]\r\n\r\n[latex]\\bar{y_{5}} = 3.6 m [\/latex]\r\n\r\n[latex]F_{R5} = 11250 N \\quad at \\quad (6.234m , 3.6 m )[\/latex]\r\n\r\n<strong>6. Review<\/strong>\r\n\r\nThis problem is less complex because of the symmetry of the cars and the bridge. The coordinates of the resultant force are just as expected. x is closer to the side with more weight (due to the distribution of weight of the vehicles), and y is less than halfway across the bridge because that side has more vehicles\r\n\r\n<\/div>\r\n<h1 id=\"chapter-1676-section-1\" class=\"section-header\">Example 3.6.9: Couple, Submitted by Anonymous<\/h1>\r\n<div class=\"textbox\">\r\n\r\n<strong>1. Problem<\/strong>\r\n<div class=\"textbox shaded\">\r\n\r\nA boy is trying to close a round-shaped water tap of diameter 1 m with a force of F = 50 N on both sides of the cylinder. Determine the couple moment produced.\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-27-e1674758566669.png\" alt=\"A 50\u202fN couple force acting 1\u202fm apart on a cylindrical object.\" width=\"180\" height=\"380\" class=\"alignnone wp-image-2044 size-full\" \/>\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\/2022\/02\/Screenshot-27-e1674758566669.png\" alt=\"A 50\u202fN couple force acting 1\u202fm apart on a cylindrical object.\" width=\"180\" height=\"380\" class=\"alignnone wp-image-2044 size-full\" \/>\r\n\r\n<strong>3. Known and Unknowns\u00a0<\/strong>\r\n\r\nKnowns:\r\n<ul>\r\n \t<li>F =50 N<\/li>\r\n \t<li>Diameter, d = 1m<\/li>\r\n<\/ul>\r\nUnknown:\r\n<ul>\r\n \t<li>Couple moments, M<\/li>\r\n<\/ul>\r\n<strong>4. Approach\u00a0<\/strong>\r\n\r\nUsing Equation [latex]M = F\\times d[\/latex]\r\n\r\n<b>5. Analysis\u00a0<\/b>\r\n\r\nCouple moment, [latex]M = F\\times d = 50N \\times 1 m= 50 Nm\u00a0 [\/latex]\r\n\r\n<strong>6. Review\u00a0\u00a0<\/strong>\r\n\r\nNote that in this question, the direction of the rotation of the tap is not mentioned. But the sign of the couple moment changes with respect to the direction of rotation and the coordinates which you choose to be positive.\r\n\r\n<\/div>","rendered":"<p>Here are examples from Chapter 3 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 3.6.1: Reaction Forces, Submitted by Andrew Williamson<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><a href=\"https:\/\/www.google.com\/url?sa=i&amp;url=https%3A%2F%2Fwww.maxpixel.net%2FSeat-Couch-Interior-Home-Furniture-Room-Sofa-42817&amp;psig=AOvVaw1_RDGb7_NnDrMzVsitaA-r&amp;ust=1629764899959000&amp;source=images&amp;cd=vfe&amp;ved=0CAoQjRxqFwoTCICj1snxxfICFQAAAAAdAAAAABAD\" rel=\"https:\/\/www.google.com\/url?sa=i&amp;url=https%3A%2F%2Fwww.maxpixel.net%2FSeat-Couch-Interior-Home-Furniture-Room-Sofa-42817&amp;psig=AOvVaw1_RDGb7_NnDrMzVsitaA-r&amp;ust=1629764899959000&amp;source=images&amp;cd=vfe&amp;ved=0CAoQjRxqFwoTCICj1snxxfICFQAAAAAdAAAAABAD\"><\/a><\/p>\n<div class=\"textbox shaded\">\n<p>A family is sitting watching TV on their couch. The couch is 5 m long and weighs 120 N. The child is sitting 1 m away from one end and has a mass of 30 kg. The mother is sitting 0.5 m away from the child and has a mass of 60 kg. The father is 3 m away from the mother and has a mass of 70 kg.<\/p>\n<p>a) Draw a free-body diagram of the couch<\/p>\n<p>b) Calculate the reaction force on each of the two legs.<\/p>\n<p>Assume the couch is supported by two rollers.<\/p>\n<figure id=\"attachment_1380\" aria-describedby=\"caption-attachment-1380\" style=\"width: 1024px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Seat-Couch-Interior-Home-Furniture-Room-Sofa-42817-1024x512.png\" alt=\"A couch\" class=\"wp-image-1380 size-large\" width=\"1024\" height=\"512\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Seat-Couch-Interior-Home-Furniture-Room-Sofa-42817-1024x512.png 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Seat-Couch-Interior-Home-Furniture-Room-Sofa-42817-300x150.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Seat-Couch-Interior-Home-Furniture-Room-Sofa-42817-768x384.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Seat-Couch-Interior-Home-Furniture-Room-Sofa-42817-65x33.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Seat-Couch-Interior-Home-Furniture-Room-Sofa-42817-225x113.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Seat-Couch-Interior-Home-Furniture-Room-Sofa-42817-350x175.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Seat-Couch-Interior-Home-Furniture-Room-Sofa-42817.png 1280w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption id=\"caption-attachment-1380\" class=\"wp-caption-text\">Source: https:\/\/www.maxpixel.net\/Seat-Couch-Interior-Home-Furniture-Room-Sofa-42817<\/figcaption><\/figure>\n<\/div>\n<hr \/>\n<p><strong>2. Draw<\/strong><\/p>\n<p>Sketch:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Andrew-1-draw-1-300x262.jpg\" alt=\"A free body diagram showing three downward forces (Fc, FM, Ff) acting on a 5\u202fm beam, with distances marked as 1\u202fm, 0.5\u202fm, and 3\u202fm between them.\" class=\"aligncenter wp-image-844 size-medium\" width=\"300\" height=\"262\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Andrew-1-draw-1-300x262.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Andrew-1-draw-1-1024x893.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Andrew-1-draw-1-768x670.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Andrew-1-draw-1-1536x1340.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Andrew-1-draw-1-65x57.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Andrew-1-draw-1-225x196.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Andrew-1-draw-1-350x305.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Andrew-1-draw-1.jpg 1600w\" 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>g = 9.81 m\/s<sup>2<\/sup><\/li>\n<li>m<sub>c<\/sub> = 30 kg<\/li>\n<li>m<sub>m<\/sub> = 60 kg<\/li>\n<li>m<sub>f<\/sub> = 70 kg<\/li>\n<li>F<sub>g<\/sub> = 120 N<\/li>\n<li>r<sub>c<\/sub> = 1 m<\/li>\n<li>r<sub>M<\/sub> = 1.5 m<\/li>\n<li>r<sub>f<\/sub> = 4.5 m<\/li>\n<li>r<sub>B<\/sub> = 5 m<\/li>\n<li>r<sub>g<\/sub> = 2.5 m<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>N<sub>A<\/sub><\/li>\n<li>N<sub>B<\/sub><\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Use equilibrium equations<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>Part a:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/andrew-1-draw-2-300x282.jpg\" alt=\"Free body diagram of a beam with applied forces (Fc, FM, Ff, Fg), reaction forces (NA, NB), and position vectors (rC, rM, rG, rF, rB) shown from point A.\" class=\"alignnone wp-image-937\" width=\"369\" height=\"347\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/andrew-1-draw-2-300x282.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/andrew-1-draw-2-1024x961.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/andrew-1-draw-2-768x721.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/andrew-1-draw-2-1536x1442.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/andrew-1-draw-2-65x61.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/andrew-1-draw-2-225x211.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/andrew-1-draw-2-350x328.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/andrew-1-draw-2.jpg 1708w\" sizes=\"auto, (max-width: 369px) 100vw, 369px\" \/><\/p>\n<p>Part b:<\/p>\n<p>[latex]\\sum M_A=0=N_B\\times r_B-F_c\\times r_c-F_M\\times r_M-F_f\\times r_f-F_g\\times r_g\\\\\\\\N_B(5m)=(30kg\\times 9.81m\/s^2)(1m)+(60kg\\times 9.81m\/s^2)(1.5m)\\\\+(70kg\\times 9.81m\/s^2)(4.5m)+(120N)(2.5m)[\/latex][latex]\\\\N_B(5m)=294.3Nm+882.9Nm +3090.15Nm+300Nm\\\\\\\\N_B(5m)=4567.35N m\\\\\\\\N_B\\frac{4567.35Nm}{5m}\\\\\\\\N_B=913.47N\\\\\\\\N_B=913N[\/latex]<\/p>\n<p>[latex]\\sum F_y=0=N_A+N_B-F_C-F_M-F_f-F_g\\\\\\\\N_A=F_C+F_M+F_f+F_g-N_B\\\\\\\\N_A=(30kg\\times 9.81m\/s^2)+(60kg\\times 9.81m\/s^2)\\\\+(70kg\\times 9.81m\/s^2)+120N-913.47N[\/latex] [latex]\\\\N_A=294.3N+588.6N+686.7N+120N-913.47N$$$$\\\\N_A=776.13N\\\\\\\\\\\\\\underline{N_A=776N}[\/latex]<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>It is interesting that N<sub>B<\/sub> is larger than N<sub>A<\/sub>, because the weight of the mother and child combined (80 kg) is larger than that of the father (70 kg). However, when you sum the moments at point B instead of A, you get the same answer. The distance between the reaction forces and the nearest forces is important, as well as the magnitude of the forces themselves. The distance between A and F<sub>c<\/sub> is 1 m, while the distance between B and F<sub>f<\/sub> is only 0.5 m.<\/p>\n<p>Additionally, it makes sense that both N<sub>A<\/sub> and N<sub>B<\/sub> are positive, i.e. are in the positive y direction.<\/p>\n<\/div>\n<h1>Example 3.6.2: Couples, Submitted by Kirsty MacLellan<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>A water valve is opened by a wheel with a diameter of 10 inches. It takes 7.5 lb of force to open the valve. What is the moment it takes to open the valve?<\/p>\n<p>Real-life scenario:<\/p>\n<figure id=\"attachment_1397\" aria-describedby=\"caption-attachment-1397\" style=\"width: 910px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/wheel-valve-heating-line.jpg\" alt=\"Three large gate valves\" class=\"wp-image-1397 size-full\" width=\"910\" height=\"640\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/wheel-valve-heating-line.jpg 910w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/wheel-valve-heating-line-300x211.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/wheel-valve-heating-line-768x540.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/wheel-valve-heating-line-65x46.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/wheel-valve-heating-line-225x158.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/wheel-valve-heating-line-350x246.jpg 350w\" sizes=\"auto, (max-width: 910px) 100vw, 910px\" \/><figcaption id=\"caption-attachment-1397\" class=\"wp-caption-text\">Source: https:\/\/www.pxfuel.com\/en\/free-photo-ekahu<\/figcaption><\/figure>\n<\/div>\n<hr \/>\n<p><strong>2. Draw<\/strong><\/p>\n<p>Sketch:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Kirsty-1-draw-1-300x281.jpg\" alt=\"10-inch wheel with opposite forces causing rotation.\" class=\"alignleft wp-image-926\" width=\"315\" height=\"295\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Kirsty-1-draw-1-300x281.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Kirsty-1-draw-1-1024x961.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Kirsty-1-draw-1-768x720.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Kirsty-1-draw-1-65x61.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Kirsty-1-draw-1-225x211.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Kirsty-1-draw-1-350x328.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Kirsty-1-draw-1.jpg 1212w\" sizes=\"auto, (max-width: 315px) 100vw, 315px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>Free-body diagram:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Kirsty-1-draw-2-1-300x224.jpg\" alt=\"A FBD of a wheel with opposing vertical forces F_A and F_B, and moment arms r_A, r_B.\" class=\"alignnone wp-image-934\" width=\"295\" height=\"220\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Kirsty-1-draw-2-1-300x224.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Kirsty-1-draw-2-1-1024x764.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Kirsty-1-draw-2-1-768x573.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Kirsty-1-draw-2-1-1536x1146.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Kirsty-1-draw-2-1-65x48.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Kirsty-1-draw-2-1-225x168.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Kirsty-1-draw-2-1-350x261.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Kirsty-1-draw-2-1.jpg 1708w\" sizes=\"auto, (max-width: 295px) 100vw, 295px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Known:<\/p>\n<ul>\n<li>d = 10 in<\/li>\n<li>F = 7.5 lb<\/li>\n<\/ul>\n<p>Unknown:<\/p>\n<ul>\n<li><span style=\"text-decoration: underline\">M<\/span><\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Determine the moment by finding the cross product of <span style=\"text-decoration: underline\">r<\/span><sub>A<\/sub> and <span style=\"text-decoration: underline\">F<\/span><sub>A<\/sub>, then <span style=\"text-decoration: underline\">r<\/span><sub>B<\/sub> and <span style=\"text-decoration: underline\">F<\/span><sub>B<\/sub>, then adding.<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>Find radius:<\/p>\n<p>[latex]r=\\frac{d}{2}\\\\r=\\frac{10in}{2}\\\\r=5in\\\\5in\\times\\frac{1ft}{12in}=0.42ft[\/latex]<\/p>\n<p>Find <span style=\"text-decoration: underline\">r<\/span><sub>A<\/sub>, <span style=\"text-decoration: underline\">F<\/span><sub>A<\/sub>, <span style=\"text-decoration: underline\">r<\/span><sub>B<\/sub>, and <span style=\"text-decoration: underline\">F<\/span><sub>B<\/sub> in vector form:<\/p>\n<p>[latex]\\underline{r}_A= \\begin{bmatrix}  0.42 \\\\  0  \\end{bmatrix}ft\\:\\; \\underline{F}_A=\\begin{bmatrix}  0 \\\\  7.5  \\end{bmatrix}lb \\\\\\underline{r}_B=\\begin{bmatrix}  -0.42 \\\\  0  \\end{bmatrix}ft\\:\\; \\underline{F}_B=\\begin{bmatrix}  0 \\\\  -7.5  \\end{bmatrix}lb[\/latex]<\/p>\n<p>Find <span style=\"text-decoration: underline\">M<\/span><sub>A<\/sub>:<\/p>\n<p>[latex]\\underline{M}_A=\\underline{r}_A\\times \\underline{F}_A=\\begin{bmatrix}  \\underline{\\hat{i}} & \\underline{\\hat{j}} & \\underline{\\hat{k}} \\\\  0.42 & 0 & 0 \\\\  0 & 7.5 & 0  \\end{bmatrix}[\/latex] [latex]\\underline{M}_A=\\hat{i} \\begin{bmatrix}  0 & 0 \\\\  7.5 & 0  \\end{bmatrix} -\\underline{\\hat{j}} \\begin{bmatrix}  0.42 & 0 \\\\  0 & 0  \\end{bmatrix}+\\underline{\\hat{k}} \\begin{bmatrix}  0.42 & 0 \\\\  0 & 7.5  \\end{bmatrix}\\\\\\underline{M}_A=(\\underline{\\hat{i}}(0)-\\underline{\\hat{j}}(0)+\\underline{\\hat{k}}(0.42\\cdot 7.5-0\\cdot 0))ft\\cdot lb\\\\\\underline{M}_A=3.15\\underline{\\hat{k}} ft\\cdot lb[\/latex]<\/p>\n<p>Find <span style=\"text-decoration: underline\">M<\/span><sub>B<\/sub>:<\/p>\n<p>[latex]\\underline{M}_B=\\underline{r}_B\\times \\underline{F}_B=\\begin{bmatrix}  \\underline{\\hat{i}} & \\underline{\\hat{j}} & \\underline{\\hat{k}} \\\\  -0.42 & 0 & 0 \\\\  0 & -7.5 & 0  \\end{bmatrix}[\/latex] [latex]\\underline{M}_B=\\hat{i} \\begin{bmatrix}  0 & 0 \\\\  -7.5 & 0  \\end{bmatrix} -\\underline{\\hat{j}} \\begin{bmatrix}  -0.42 & 0 \\\\  0 & 0  \\end{bmatrix}+\\underline{\\hat{k}} \\begin{bmatrix}  -0.42 & 0 \\\\  0 & -7.5  \\end{bmatrix}\\\\\\underline{M}_B=(\\underline{\\hat{i}}(0)-\\hat{j}(0)+\\underline{\\hat{k}}(-0.42\\cdot -7.5-0\\cdot 0))ft\\cdot lb\\\\\\underline{M}_B=3.15\\underline{\\hat{k}} ft\\cdot lb[\/latex]<\/p>\n<p>Add <span style=\"text-decoration: underline\">M<\/span><sub>A<\/sub> and <span style=\"text-decoration: underline\">M<\/span><sub>B<\/sub> to get <span style=\"text-decoration: underline\">M<\/span>:<\/p>\n<p>[latex]\\underline{M}=\\underline{M}_A+\\underline{M}_B\\\\\\underline{M}=3.15ft\\cdot lb+3.15ft\\cdot lb\\\\\\underline{M}=6.3\\underline{\\hat{k}}ft\\cdot lb[\/latex]<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>Notice that for the x-coordinates, the positive x-direction was taken to the left, and this was consistently followed in the calculation. It&#8217;s important to stay consistent with your chosen direction, as mixing it up can lead to sign errors.<\/p>\n<p>This answer makes sense because there is only one moment acting in the k direction.<\/p>\n<p>Note: We could have come to the same answer using the formula M = F*d, which would have been faster.<\/p>\n<p>[latex]M=f\\cdot d\\\\M=7.5lb\\cdot 10in(\\frac{1ft}{12in})\\\\M=7.5lb\\cdot \\frac{5}{6}ft\\\\\\\\M=6.25ft\\cdot lb[\/latex]<\/p>\n<p>This answer is slightly more accurate because we didn&#8217;t round when converting between inches and feet (in the original solution, we rounded 0.416667 to 0.42).<\/p>\n<\/div>\n<h1>Example 3.6.3: Distributed Load, Submitted by Luciana Davila<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>A shelf on the wall is 1.5 meters away from the floor. The shelf has a length of 100 cm. A person starts putting different objects on it to create a distributed load. The load created a curve described by:<\/p>\n<p style=\"text-align: center\">w = 4x<sup>4<\/sup> +2 N\/m.<\/p>\n<p>Calculate the resultant force and how far it is acting from the wall (Fixed end).<\/p>\n<figure style=\"width: 300px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/shelf-576088_1280-300x150.png\" alt=\"A shelf\" width=\"300\" height=\"150\" style=\"border: 1px solid black\" \/><figcaption class=\"wp-caption-text\">Source: https:\/\/pixabay.com\/vectors\/shelf-floating-bathroom-glass-576088\/<\/figcaption><\/figure>\n<\/div>\n<hr \/>\n<p><strong>2. Draw<\/strong><\/p>\n<p>Sketch:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luciana-1-draw-2-275x300.jpg\" alt=\"A triangular distributed load on a cantilever beam, 100\u202fcm long and 1.5\u202fm high.\" class=\"alignleft wp-image-945 size-medium\" width=\"275\" height=\"300\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luciana-1-draw-2-275x300.jpg 275w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luciana-1-draw-2-939x1024.jpg 939w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luciana-1-draw-2-768x837.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luciana-1-draw-2-1409x1536.jpg 1409w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luciana-1-draw-2-65x71.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luciana-1-draw-2-225x245.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luciana-1-draw-2-350x382.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luciana-1-draw-2.jpg 1437w\" sizes=\"auto, (max-width: 275px) 100vw, 275px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>Free-body diagram:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luciana-1-draw-1-2-300x156.jpg\" alt=\"Free body diagram of a cantilever beam with a variable load w = 4x^4 + 2, showing reaction forces and moment at the fixed end.\" class=\"alignnone wp-image-947\" width=\"440\" height=\"229\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luciana-1-draw-1-2-300x156.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luciana-1-draw-1-2-1024x531.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luciana-1-draw-1-2-768x398.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luciana-1-draw-1-2-1536x797.jpg 1536w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luciana-1-draw-1-2-2048x1063.jpg 2048w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luciana-1-draw-1-2-65x34.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luciana-1-draw-1-2-225x117.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/07\/Luciana-1-draw-1-2-350x182.jpg 350w\" sizes=\"auto, (max-width: 440px) 100vw, 440px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>w = 4x<sup>4<\/sup> + 2 N\/m<\/li>\n<li>L = 100 cm = 1m<\/li>\n<li>xmin = 0m<\/li>\n<li>xmax = 1m<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>x<sub>r<\/sub><\/li>\n<li>F<sub>r<\/sub><\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Use distributed load equations:<\/p>\n<p>[latex]F_r=\\int^{xmax}_{xmin} wdx\\\\X_r=\\frac{\\int^{xmax}_{xmin} x*w(x)*dx}{\\int^{xmax}_{xmin}wdx}[\/latex]<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>Solve for F<sub>r<\/sub>:<\/p>\n<p>[latex]F_r=\\int^1_0 (4x^4+2)dx\\;\\; N\\\\F_r=(\\frac{4x^5}{5}+2x)\\vert^1_0\\;\\;N\\\\F_r=(\\frac{4}{5}+2)N\\\\F_r=2.8N[\/latex]<\/p>\n<p>Solve for x<sub>r<\/sub>:<\/p>\n<p>[latex]X_r=\\frac{(\\int^1_0x(4x^4+2)dx)N\/m}{\\int^1_0(4x^4+2)dx)N}\\\\X_r=\\frac{\\int^1_0(4x^5+2x)dxN\/m}{2.8N}\\\\X_r=\\frac{(\\frac{4x^6}{6}+\\frac{2x^2}{2})\\vert^1_0N\/m}{2.8N}\\\\X_r=\\frac{(\\frac{2}{3}+1)N\/m}{2.8N}\\\\X_r=0.59m[\/latex]<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>The function shows an increasing curve on the interval, so it makes sense that the resultant force would be applied closer to the right end of the beam than the left end.<\/p>\n<\/div>\n<h1>Example 3.6.4: Couple, Submitted by Hamza <span>Ben Driouech<\/span><\/h1>\n<div class=\"textbox\">\n<ol>\n<li><strong>Problem<\/strong><\/li>\n<\/ol>\n<div class=\"textbox shaded\">\n<p>A mason jar lid with a diameter of 11cm is firmly closed and requires a moment of 5.5 Nm to be opened. If you are to open the lid with your finger and thumb, assuming they are applying equal force, determine the force required by your thumb and finger.<\/p>\n<figure id=\"attachment_1847\" aria-describedby=\"caption-attachment-1847\" style=\"width: 409px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/jar-lid-opening.jpg\" alt=\"A person tightly gripping and twisting the lid of a jar to open it.\" class=\"wp-image-1847\" width=\"409\" height=\"318\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/jar-lid-opening.jpg 255w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/jar-lid-opening-65x50.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/jar-lid-opening-225x175.jpg 225w\" sizes=\"auto, (max-width: 409px) 100vw, 409px\" \/><figcaption id=\"caption-attachment-1847\" class=\"wp-caption-text\">Source: https:\/\/commons.wikimedia.org\/wiki\/File:Fingertip_tightening_lid_on_pickled_beet_jar.jpg<\/figcaption><\/figure>\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\/2022\/02\/Screenshot-2025-05-07-140431-223x300.png\" alt=\"A free-body diagram of a jar lid showing a 5.5 Nm moment.\" width=\"223\" height=\"300\" style=\"border: 1px solid black\" class=\"alignnone\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>M = 5.5 Nm<\/li>\n<li>d = 11 cm = 0.11 m<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>F &#8211; Force exerted by thumb and finger<\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Using couple to find F<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>Since the forces exerted by the finger and thumb are equal and opposite, it is possible to find the force exerted by the finger and thumb using the couple equation.<\/p>\n<p>[latex]5.5Nm=F\\times0.11m[\/latex]<\/p>\n<p>[latex]F=\\frac{5.5Nm}{0.11m}\\\\F =50N[\/latex]<\/p>\n<p>Thus, your thumb and finger are applying a force of 50N each.<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>The moment equation is applicable since the force of the thumb and finger is equal. The equation produces a reasonable value of 50N<\/p>\n<\/div>\n<h1>Example 3.6.5: Couple, Submitted by Andrew Williamson.<\/h1>\n<div class=\"textbox\">\n<ol>\n<li><strong>Problem<\/strong><\/li>\n<\/ol>\n<div class=\"textbox shaded\">\n<p>You are driving home from work, and you need to turn right to get into your driveway. To turn the steering, you are applying 115 N with each hand. Consider your steering wheel has a diameter of 40 cm.<\/p>\n<p>a. Determine the couple moment produced on the steering wheel.<\/p>\n<p>b. Compare the above result with a situation where you use single hand to navigate the wheel. How much force does it take to turn the steering wheel in this case?<\/p>\n<figure id=\"attachment_1836\" aria-describedby=\"caption-attachment-1836\" style=\"width: 404px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Andrew-williamson-wheel-problem.jpg\" alt=\"A person turning a steering wheel\" class=\"wp-image-1836\" width=\"404\" height=\"269\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Andrew-williamson-wheel-problem.jpg 275w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Andrew-williamson-wheel-problem-65x43.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Andrew-williamson-wheel-problem-225x150.jpg 225w\" sizes=\"auto, (max-width: 404px) 100vw, 404px\" \/><figcaption id=\"caption-attachment-1836\" class=\"wp-caption-text\">Source:https:\/\/commons.wikimedia.org\/wiki\/File:Woman_hand_steering_wheel_%28Unsplash%29.jpg<\/figcaption><\/figure>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/hands-on-wheel-andrew-williamson-e1652710940105.jpg\" alt=\"A sketch of a person turning the steering wheel.\" class=\"alignnone wp-image-1837\" width=\"326\" height=\"271\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/hands-on-wheel-andrew-williamson-e1652710940105.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/hands-on-wheel-andrew-williamson-e1652710940105-65x54.jpg 65w\" sizes=\"auto, (max-width: 326px) 100vw, 326px\" \/><\/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\/andrew-williamson-steeering-wheel-6.jpg\" alt=\"Two steering wheel setups: one with opposite 115\u202fN forces creating a moment, the other with one angled force F_2.\" class=\"alignnone wp-image-1846 size-full\" width=\"701\" height=\"334\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/andrew-williamson-steeering-wheel-6.jpg 701w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/andrew-williamson-steeering-wheel-6-300x143.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/andrew-williamson-steeering-wheel-6-65x31.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/andrew-williamson-steeering-wheel-6-225x107.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/andrew-williamson-steeering-wheel-6-350x167.jpg 350w\" sizes=\"auto, (max-width: 701px) 100vw, 701px\" \/><\/p>\n<p><strong><span style=\"text-align: initial;font-size: 1em\">3. Knowns and Unknowns<\/span><\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>F = 115 N<\/li>\n<li>d = 40 cm = 0.40m<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>Couple moment, M<\/li>\n<li>Force applied by hand in the second scenario, F<sub>2<\/sub><\/li>\n<\/ul>\n<p><strong>4. Approach<\/strong><\/p>\n<p>Use the couple Moment equation to find the moment, and then use the moment to find the force required to turn the handle single-handedly.<\/p>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>a.\u00a0 Both arms are applying a 115 N force each. The diameter of the wheel is 40 cm.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/hands-on-wheel-andrew-williamson-e1652710940105.jpg\" alt=\"A sketch of a person turning the steering wheel.\" class=\"alignnone wp-image-1837 size-full\" width=\"225\" height=\"187\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/hands-on-wheel-andrew-williamson-e1652710940105.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/hands-on-wheel-andrew-williamson-e1652710940105-65x54.jpg 65w\" sizes=\"auto, (max-width: 225px) 100vw, 225px\" \/><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/andrew-williamson-steeering-wheel-2.jpg\" alt=\"Two opposite 115\u202fN forces act on a 0.4\u202fm diameter wheel to create a moment.\" class=\"alignnone wp-image-1840 size-full\" width=\"306\" height=\"293\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/andrew-williamson-steeering-wheel-2.jpg 306w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/andrew-williamson-steeering-wheel-2-300x287.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/andrew-williamson-steeering-wheel-2-65x62.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/andrew-williamson-steeering-wheel-2-225x215.jpg 225w\" sizes=\"auto, (max-width: 306px) 100vw, 306px\" \/><\/p>\n<p>Therefore,<\/p>\n<p>[latex]\\overrightarrow{M}= \\overrightarrow{F}\\times\\overrightarrow{d}=115N\\times0.4m[\/latex]<\/p>\n<p>[latex]\\overrightarrow{M}=46Nm[\/latex]<\/p>\n<p>b. Consider 46Nm as the required moment to turn the steering wheel.<\/p>\n<p>When just one hand applies force on the wheel it forms a moment which is equal to 46 Nm, with distance and angle given we can find the force it takes to turn the steering wheel.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/andrew-williamson-steeering-wheel-4.jpg\" alt=\"Force F_2 is applied to a 0.4\u202fm diameter wheel to produce a moment.\" class=\"aligncenter wp-image-1842\" width=\"288\" height=\"249\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/andrew-williamson-steeering-wheel-4.jpg 256w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/andrew-williamson-steeering-wheel-4-65x56.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/andrew-williamson-steeering-wheel-4-225x194.jpg 225w\" sizes=\"auto, (max-width: 288px) 100vw, 288px\" \/><\/p>\n<p>[latex]\\overrightarrow{M} = F_{2}\\times\\frac{d}{2}\\times\\sin{\\theta}\\\\46Nm =F_{2}\\cdot\\frac{0.4m}{2}\\cdot\\sin{90}\\\\F_{2}=230 N[\/latex]<\/p>\n<p><strong>6. Review\u00a0<\/strong><\/p>\n<p>As expected, the force it took to turn the steering wheel is twice the force it took with one hand when both hands are used.<\/p>\n<\/div>\n<h1>Example 3.6.6: Couple, Submitted by Elliot Fraser<\/h1>\n<div class=\"textbox\">\n<ol>\n<li><strong>Problem<\/strong><\/li>\n<\/ol>\n<div class=\"textbox shaded\">\n<p>Ellen is taking her driving test and is asked to perform a right turn. The diameter of the steering wheel is 15 inches. Ellen&#8217;s right hand exerts a force of 200 N, and her left exerts 150 N.<\/p>\n<p>a. Calculate the net moment in the above-given situation using the cross product.<\/p>\n<p>b. If Ellen were to perform a left turn with the same forces given, what would be the net moment?<\/p>\n<figure id=\"attachment_1901\" aria-describedby=\"caption-attachment-1901\" style=\"width: 705px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Elliot-fraser-steering-wheel.jpg\" alt=\"A person turning a Toyota steering wheel with both hands.\" class=\"wp-image-1901\" width=\"705\" height=\"470\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Elliot-fraser-steering-wheel.jpg 1200w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Elliot-fraser-steering-wheel-300x200.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Elliot-fraser-steering-wheel-1024x683.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Elliot-fraser-steering-wheel-768x512.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Elliot-fraser-steering-wheel-65x43.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Elliot-fraser-steering-wheel-225x150.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Elliot-fraser-steering-wheel-350x233.jpg 350w\" sizes=\"auto, (max-width: 705px) 100vw, 705px\" \/><figcaption id=\"caption-attachment-1901\" class=\"wp-caption-text\">Source:https:\/\/www.wikihow.com\/Drive-Safely-in-Fog<\/figcaption><\/figure>\n<p>The above image is a close illustration of the problem. In the problem, the hands of the driver are along the diameter.<\/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\/2022\/02\/Elliot-fraser-chap-3-1.jpg\" alt=\"Diagram showing right and left steering turns with opposing forces applied 15 inches apart.\" class=\"alignnone wp-image-1905 size-full\" width=\"726\" height=\"426\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Elliot-fraser-chap-3-1.jpg 726w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Elliot-fraser-chap-3-1-300x176.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Elliot-fraser-chap-3-1-65x38.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Elliot-fraser-chap-3-1-225x132.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Elliot-fraser-chap-3-1-350x205.jpg 350w\" sizes=\"auto, (max-width: 726px) 100vw, 726px\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>Diameter of the wheel = 15 in = 0.381m<\/li>\n<li>Right hand force = 200 N<\/li>\n<li>left hand force = 150 N<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>M<sub>1\u00a0<\/sub>(for right hand)<\/li>\n<li>M<sub>2 <\/sub>(for left hand)<\/li>\n<\/ul>\n<p><strong>4. Approach\u00a0<\/strong><\/p>\n<p>Convert inches to meters and add the cross product of distance and force to find the net moment.<\/p>\n<p><strong>5. Analysis\u00a0<\/strong><\/p>\n<p>a.RIGHT TURN<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-21.png\" alt=\"A diagram of a right turn showing a couple: 150\u202fN upward and 200\u202fN downward, 15\u202fin apart.\" class=\"alignnone wp-image-1912 size-full\" width=\"383\" height=\"414\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-21.png 383w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-21-278x300.png 278w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-21-65x70.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-21-225x243.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-21-350x378.png 350w\" sizes=\"auto, (max-width: 383px) 100vw, 383px\" \/><\/p>\n<p>Express the radius of the steering wheel and forces in vector form for tight turn.<\/p>\n<p>[latex]\\overrightarrow {r_{1}} = \\begin{bmatrix}-0.1905\\\\0\\\\0\\end{bmatrix}\\kern 1pc\\overrightarrow {r_{2}} = \\begin{bmatrix}0.1905\\\\0\\\\0\\end{bmatrix}[\/latex]<\/p>\n<p>[latex]\\overrightarrow {F_{1}} = \\begin{bmatrix}0\\\\150\\\\0\\end{bmatrix}\\kern 1pc\\overrightarrow {F_{2}} = \\begin{bmatrix}0\\\\-200\\\\0\\end{bmatrix}[\/latex]<\/p>\n<p>[latex]\\overrightarrow{M_1}=\\overrightarrow{r_{1}} \\times \\overrightarrow{F_{1}} \\kern 1pc +\u00a0 \\kern 1pc\\overrightarrow{r_{2}} \\times \\overrightarrow{F_{2}}[\/latex]<\/p>\n<p><span style=\"text-align: initial;font-size: 1em\">[latex]\\overrightarrow{M_{1}}=\\:\\begin{vmatrix} \\hat{i} & \\hat{j} & \\hat{k} \\\\-0.1905 & 0 & 0\\\\0 &\u00a0 150 & 0\\end{vmatrix}[\/latex][latex]\\kern 1pc + \\kern 1pc\u00a0 \\begin{vmatrix} \\hat{i} & \\hat{j} & \\hat{k} \\\\0.1905 & 0 & 0\\\\0 &\u00a0 -200\u00a0 & 0\\end{vmatrix}[\/latex][latex]\\:=-28.575\\: \\hat{k} - 38.1 \\hat{k}[\/latex]<\/span><\/p>\n<p>[latex]\\overrightarrow{M_{1}}=\\: -66.675 \\: \\hat{k}[\/latex]<\/p>\n<p>b. LEFT TURN<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-22.png\" alt=\"A diagram of a left turn showing a couple: 200\u202fN upward and 150\u202fN downward, 15\u202fin apart.\" class=\"alignnone wp-image-1913 size-full\" width=\"361\" height=\"437\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-22.png 361w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-22-248x300.png 248w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-22-65x79.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-22-225x272.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-22-350x424.png 350w\" sizes=\"auto, (max-width: 361px) 100vw, 361px\" \/><\/p>\n<p>Radius and force in vector form<\/p>\n<p>Radius is the same as it is for a right turn, but the forces have a different direction.<\/p>\n<p>[latex]\\overrightarrow {r_{1}} = \\begin{bmatrix}-0.1905\\\\0\\\\0\\end{bmatrix}\\kern 1pc\\overrightarrow {r_{2}} = \\begin{bmatrix}0.1905\\\\0\\\\0\\end{bmatrix}[\/latex]<\/p>\n<p>[latex]\\overrightarrow {F_{3}} = \\begin{bmatrix}0\\\\-150\\\\0\\end{bmatrix}\\kern 1pc\\overrightarrow {F_{4}} = \\begin{bmatrix}0\\\\200\\\\0\\end{bmatrix}[\/latex]<\/p>\n<p>[latex]\\overrightarrow{M_2}=\\overrightarrow{r_{1}} \\times \\overrightarrow{F_{3}} \\kern 1pc +\u00a0 \\kern 1pc\\overrightarrow{r_{2}} \\times \\overrightarrow{F_{4}}[\/latex]<\/p>\n<p><span style=\"text-align: initial;font-size: 1em\">[latex]\\overrightarrow{M_{2}}=\\:\\begin{vmatrix} \\hat{i} & \\hat{j} & \\hat{k} \\\\-0.1905 & 0 & 0\\\\0 &\u00a0 -150 & 0\\end{vmatrix}[\/latex][latex]\\kern 1pc + \\kern 1pc\u00a0 \\begin{vmatrix} \\hat{i} & \\hat{j} & \\hat{k} \\\\0.1905 & 0 & 0\\\\0 &\u00a0 200\u00a0 & 0\\end{vmatrix}[\/latex][latex]\\:=28.575\\: \\hat{k} + 38.1 \\hat{k}[\/latex]<\/span><\/p>\n<p>&nbsp;<\/p>\n<p>[latex]\\overrightarrow{M_{2}}=\\: 66.675 \\: \\hat{k}[\/latex]<\/p>\n<p><strong>6. Review\u00a0<\/strong><\/p>\n<p>Because the radius and force values in both instances have the same magnitude but different directions, the net moments for left and right turns were expected to have the same magnitudes and different directions; thus, the answers make sense.<\/p>\n<\/div>\n<h1 id=\"chapter-1676-section-1\" class=\"section-header\">Example 3.6.7: Lever Arms, Submitted by Dhruvil Kanani<\/h1>\n<div class=\"textbox\">\n<p>1. <strong>Problem<\/strong><\/p>\n<div class=\"textbox shaded\">Josh has two sons; the older one weighs 50 kg, while the younger one weighs 35 kg. Josh wants to<br \/>\nbuild a Seesaw (tatter-totter) for his sons in a way that both the kids have to apply an equal amount<br \/>\nof force. What is the required ratio of the lengths of both arms to achieve this? Acceleration due to gravity is 9.81m\/s<sup>2<\/sup>.<\/div>\n<p><strong>2. Sketch:<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Chapter3_Dhru_1-1-300x136.png\" alt=\"A simply supported beam with loads applied at different positions along its length.\" width=\"300\" height=\"136\" class=\"alignnone wp-image-2120 size-medium\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Chapter3_Dhru_1-1-300x136.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Chapter3_Dhru_1-1-65x30.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Chapter3_Dhru_1-1-225x102.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Chapter3_Dhru_1-1-350x159.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Chapter3_Dhru_1-1.png 622w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Chapter3_Dhru_2-1-300x125.png\" alt=\"A free body diagram of the problem.\" width=\"300\" height=\"125\" class=\"alignnone wp-image-2121 size-medium\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Chapter3_Dhru_2-1-300x125.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Chapter3_Dhru_2-1-65x27.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Chapter3_Dhru_2-1-225x94.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Chapter3_Dhru_2-1-350x146.png 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Chapter3_Dhru_2-1.png 581w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><br \/>\n<strong>3. Knowns and Unknowns:<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li><span>m<\/span><sub>1<\/sub><span>\u00a0= 35kg<\/span><\/li>\n<li><span>m<\/span><sub>2<\/sub><span> = 50kg<\/span><\/li>\n<li><span>g = 9.81 m\/s<\/span><sup>2<\/sup><\/li>\n<\/ul>\n<p>Unknown:<\/p>\n<ul>\n<li><span>R<\/span><sub>1<\/sub><\/li>\n<li><span>R<\/span><sub>2<\/sub><\/li>\n<\/ul>\n<p><strong style=\"text-align: initial;background-color: initial;font-size: 1em\">4. Approach:<\/strong><\/p>\n<p>Convert the masses to forces using gravity<strong>.<br \/>\n<\/strong>Equate the moments caused by each force.<br \/>\nSolve for the ratio of lengths.<\/p>\n<p><strong>5. Analysis:<\/strong><br \/>\nFor the children to apply an equal amount of force while playing on the seesaw, they must create an equal moment.<br \/>\nTherefore:<br \/>\n[latex]M_1 = M_2[\/latex]<br \/>\nThese moments can be rewritten in terms of F and R.<br \/>\n[latex]F_1 \\cdot R_1 = F_2 \\cdot R_2[\/latex]<br \/>\nThe F values can be solved by multiplying the child&#8217;s mass by gravity to find the force they exert on the seesaw.<br \/>\n[latex](m_1 \\cdot g) \\cdot R_1 = (m_2 \\cdot g) \\cdot R_2[\/latex]<br \/>\nGravity can be cancelled out as it exists on both sides of the equation, leaving:<br \/>\n[latex]m_1\u00a0 \\cdot R_1 = m_2 \\cdot R_2[\/latex]<br \/>\nPutting values into this equation gives:<br \/>\n[latex]35kg\u00a0 \\cdot R_1 = 50kg \\cdot R_2[\/latex]<br \/>\nThe equation can now be adjusted to solve for R<sub>1<\/sub>\/R<sub>2<\/sub>.<br \/>\n[latex]\\frac{R_1}{R_2} = \\frac{50kg}{35kg}[\/latex]<br \/>\nThis can be simplified to give the ratio:<br \/>\n[latex]\\frac{R_1}{R_2} = \\frac{10}{7} \\\\ R_1 : R_2 = 10 : 7[\/latex]<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>It makes sense that the lighter child needs to be further away to produce an equal moment to that of the heavier child. 10:7 is a reasonable ratio that matches the ratio of their weights.<\/p>\n<\/div>\n<h1 class=\"section-header\">Example 3.6.8: Distributed loads, Submitted by Will Craine<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>There are 5 cars on a bridge, three in one lane and the rest in the second lane. Each car is of identical dimensions and weight.<\/p>\n<p>a. Find the resultant force of the cars from the side of the bridge with 3 cars.<\/p>\n<p>b. Find the resultant force of the cars from the side of the bridge, with 2 cars in the opposite direction to the other 3 cars.<\/p>\n<p>c. Find the resultant force of all 5 cars and the location of this force using the average of the x and y coordinates.<\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Updates-car-300x163.jpg\" alt=\"Layout of objects A to E in a 13\u202fm wide area.\" width=\"491\" height=\"267\" class=\"aligncenter wp-image-2125\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Updates-car-300x163.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Updates-car-65x35.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Updates-car-225x122.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Updates-car-350x190.jpg 350w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Updates-car.jpg 748w\" sizes=\"auto, (max-width: 491px) 100vw, 491px\" \/><span style=\"text-align: initial;font-size: 1em\">Figure 1: Top view of the bridge\u00a0<\/span><\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-20-1-e1674753024654.png\" alt=\"A 3D load layout over a 13\u202fm span, and 1000\u202fN\/m triangular distributed loads.\" width=\"574\" height=\"386\" class=\"aligncenter wp-image-2033\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-20-1-e1674753024654.png 840w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-20-1-e1674753024654-300x202.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-20-1-e1674753024654-768x517.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-20-1-e1674753024654-65x44.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-20-1-e1674753024654-225x151.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-20-1-e1674753024654-350x235.png 350w\" sizes=\"auto, (max-width: 574px) 100vw, 574px\" \/><\/p>\n<p style=\"text-align: center\">Figure 2: Drawing denoting direction and dimension of cars and bridge (Assuming the cars are in a Trapezium Shape).<\/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\/2022\/02\/Screenshot-20-1-e1674753024654.png\" alt=\"A 3D load layout over a 13\u202fm span, and 1000\u202fN\/m triangular distributed loads.\" width=\"479\" height=\"322\" class=\"alignnone wp-image-2033\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-20-1-e1674753024654.png 840w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-20-1-e1674753024654-300x202.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-20-1-e1674753024654-768x517.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-20-1-e1674753024654-65x44.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-20-1-e1674753024654-225x151.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-20-1-e1674753024654-350x235.png 350w\" sizes=\"auto, (max-width: 479px) 100vw, 479px\" \/><\/p>\n<p><strong>3. Knowns and Unknowns<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>Dimensions of the cars and the bridge<\/li>\n<li>Weight of one car, w = 1000 N\/m<\/li>\n<\/ul>\n<p>Unknowns:<\/p>\n<ul>\n<li>The resultant force on either side of the bridge.<\/li>\n<li>The location of the resultant force of all 5 cars.<\/li>\n<\/ul>\n<p><strong>4.\u00a0 Approach<\/strong><\/p>\n<ul>\n<li>The weight of the cars is distributed along the dimensions. We figure out he resultant forces and the location of the resultant forces along the x direction using the equation [latex]\\bar{x}=\\frac{\\sum F_{i}x_i}{\\sum F_i}[\/latex].<\/li>\n<li>Note that the shape of the car consists of a rectangle and an adjacent triangle. Hence, the distributed load is found for both shapes, using equations appropriate to find the area under the force, and then added to form the resultant force of one car.<\/li>\n<\/ul>\n<p><strong>5. Analysis<\/strong><\/p>\n<p>a. <span style=\"text-decoration: underline\">Finding the location along the x-axis and the magnitude of the resultant force of 3 cars<\/span><\/p>\n<p>Since all 3 cars are identical [latex]F_{A}=F_{B}=F_{C}[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-21-1-e1674754059542.png\" alt=\"Triangular uniform distributed loads over a 3\u202fm span with 1000\u202fN\/m.\" width=\"604\" height=\"232\" class=\"alignnone wp-image-2036 size-full\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-21-1-e1674754059542.png 604w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-21-1-e1674754059542-300x115.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-21-1-e1674754059542-65x25.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-21-1-e1674754059542-225x86.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-21-1-e1674754059542-350x134.png 350w\" sizes=\"auto, (max-width: 604px) 100vw, 604px\" \/><\/p>\n<p>From the fig above<\/p>\n<p>[latex]F_{R1}=1000 N\/m \\times 1.5 m = 1500 N[\/latex]<\/p>\n<p>[latex]F_{R2} = \\frac{1}{2} \\times 1.5 m \\times 1000 N\/m = 750 N[\/latex]<\/p>\n<p>[latex]F_{A} = 1500 N + 750 N = 2250N[\/latex]<\/p>\n<p><span style=\"text-decoration: underline\">[latex]\\bar {x}[\/latex] for resultant force of car A<\/span><\/p>\n<p>&nbsp;<\/p>\n<p>From O [latex]F_{R1}[\/latex] locates at [latex]1+ \\frac{1}{2}(1.5) = 1.75 m[\/latex]<\/p>\n<p>[latex]F_{R2}[\/latex] locates at [latex]1+ 1.5+ \\frac{1}{3}(1.5) = 3 m[\/latex]<\/p>\n<p>now, [latex]\\bar{x}=\\frac{\\sum F_{i}x_i}{\\sum F_i}[\/latex]<\/p>\n<p>thus [latex]\\bar{x_{A}}= \\frac {(1500 N\/m\\times 1.75m+(750 N\/m\\times 3m)}{(1500N +750N)}[\/latex]<\/p>\n<p>[latex]\\bar{x_{A}} = 2.17 m[\/latex] from left side.<\/p>\n<p>Also for one car the [latex]\\bar{x}[\/latex] of 2250 N force is [latex]2.17m - 1 m = 1.17 m[\/latex]<\/p>\n<p>Using these calculations, the following figure can be drawn.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-24-e1674755299302.png\" alt=\"Locations of forces A, B, and C with their x-coordinates.\" width=\"800\" height=\"216\" class=\"alignnone wp-image-2039 size-full\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-24-e1674755299302.png 800w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-24-e1674755299302-300x81.png 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-24-e1674755299302-768x207.png 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-24-e1674755299302-65x18.png 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-24-e1674755299302-225x61.png 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-24-e1674755299302-350x95.png 350w\" sizes=\"auto, (max-width: 800px) 100vw, 800px\" \/><\/p>\n<p>The total resultant force of all 3 cars is [latex]F_{R3} = 2250N \\times 3 = 6750 N[\/latex]<\/p>\n<p>[latex]\\bar{x}=\\frac{\\sum F_{i}x_i}{\\sum F_i}[\/latex]<\/p>\n<p>therefore,<\/p>\n<p>[latex]\\bar{x} = \\frac{(2.17 \\times 2250)+(6.17\\times 2250)+(10.17\\times 2250)}{6750}[\/latex]<\/p>\n<p>[latex]\\bar{x} = 6.17 m[\/latex]\u00a0 from left side .<\/p>\n<p>&nbsp;<\/p>\n<p>b. <span style=\"text-decoration: underline\">Position and magnitude of the resultant force of the 2 cars on the lane.\u00a0<\/span><\/p>\n<p>[latex]F_{D} = F_{E}=2250 N[\/latex]<\/p>\n<p>Total force of the 2 cars [latex]F_{R2} = 2250 \\times 2= 4500 N[\/latex]<\/p>\n<p>Position of the resultant forces from the 2 cars,<\/p>\n<p>[latex]\\bar {x_{2}} = \\frac {(4.17 \\times 2250) + (9.17 \\times 2250)}{4500}=6.67 m[\/latex]<\/p>\n<p>6.67 m is from the right side, but from the\u00a0 left side\u00a0 [latex]\\bar{x_{2}} = 13-6.67 = 6.33 m[\/latex]<\/p>\n<p>c. Position and magnitude of the total resultant force on the bridge.<\/p>\n<p>Total force on bridge due to the 5 cars [latex]F_{R5} =F_{R2} +F_{R3} = 4500 N + 6750 N = 11250 N[\/latex]<\/p>\n<p>[latex]\\bar{x} = \\frac{(F_{R2}\\times \\bar{x_{2}})+(F_{R3}\\times\\bar{x_{3}})}{F_{R5}}[\/latex]<\/p>\n<p>[latex]\\bar {x} =\\frac{ (6.33 \\times 4500)+ (6.17 \\times 6750 )}{11250}= 6.234 m[\/latex]<\/p>\n<p>[latex]\\bar {y}\u00a0[\/latex] for all the cars is half of its width<\/p>\n<p>therefore, [latex]\\bar{y_{5}} = \\frac{(\\bar{y_{2}}\\times F_{R3} )+(\\bar{}y_{3}\\ times F_{R3})}{F_{R5}}[\/latex]<\/p>\n<p>[latex]\\bar{y_{5}} = \\frac{(6m \\times 4500 N )+(2m \\times 6750 N)}{11250 N}[\/latex]<\/p>\n<p>[latex]\\bar{y_{5}} = 3.6 m[\/latex]<\/p>\n<p>[latex]F_{R5} = 11250 N \\quad at \\quad (6.234m , 3.6 m )[\/latex]<\/p>\n<p><strong>6. Review<\/strong><\/p>\n<p>This problem is less complex because of the symmetry of the cars and the bridge. The coordinates of the resultant force are just as expected. x is closer to the side with more weight (due to the distribution of weight of the vehicles), and y is less than halfway across the bridge because that side has more vehicles<\/p>\n<\/div>\n<h1 class=\"section-header\">Example 3.6.9: Couple, Submitted by Anonymous<\/h1>\n<div class=\"textbox\">\n<p><strong>1. Problem<\/strong><\/p>\n<div class=\"textbox shaded\">\n<p>A boy is trying to close a round-shaped water tap of diameter 1 m with a force of F = 50 N on both sides of the cylinder. Determine the couple moment produced.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-27-e1674758566669.png\" alt=\"A 50\u202fN couple force acting 1\u202fm apart on a cylindrical object.\" width=\"180\" height=\"380\" class=\"alignnone wp-image-2044 size-full\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-27-e1674758566669.png 180w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-27-e1674758566669-142x300.png 142w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-27-e1674758566669-65x137.png 65w\" sizes=\"auto, (max-width: 180px) 100vw, 180px\" \/><\/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\/2022\/02\/Screenshot-27-e1674758566669.png\" alt=\"A 50\u202fN couple force acting 1\u202fm apart on a cylindrical object.\" width=\"180\" height=\"380\" class=\"alignnone wp-image-2044 size-full\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-27-e1674758566669.png 180w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-27-e1674758566669-142x300.png 142w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2022\/02\/Screenshot-27-e1674758566669-65x137.png 65w\" sizes=\"auto, (max-width: 180px) 100vw, 180px\" \/><\/p>\n<p><strong>3. Known and Unknowns\u00a0<\/strong><\/p>\n<p>Knowns:<\/p>\n<ul>\n<li>F =50 N<\/li>\n<li>Diameter, d = 1m<\/li>\n<\/ul>\n<p>Unknown:<\/p>\n<ul>\n<li>Couple moments, M<\/li>\n<\/ul>\n<p><strong>4. Approach\u00a0<\/strong><\/p>\n<p>Using Equation [latex]M = F\\times d[\/latex]<\/p>\n<p><b>5. Analysis\u00a0<\/b><\/p>\n<p>Couple moment, [latex]M = F\\times d = 50N \\times 1 m= 50 Nm\u00a0[\/latex]<\/p>\n<p><strong>6. Review\u00a0\u00a0<\/strong><\/p>\n<p>Note that in this question, the direction of the rotation of the tap is not mentioned. But the sign of the couple moment changes with respect to the direction of rotation and the coordinates which you choose to be positive.<\/p>\n<\/div>\n","protected":false},"author":60,"menu_order":6,"template":"","meta":{"pb_show_title":"on","pb_short_title":"3.6 Examples","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-569","chapter","type-chapter","status-publish","hentry"],"part":54,"_links":{"self":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/569","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":58,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/569\/revisions"}],"predecessor-version":[{"id":2850,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/569\/revisions\/2850"}],"part":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/parts\/54"}],"metadata":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/569\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/media?parent=569"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapter-type?post=569"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/contributor?post=569"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/license?post=569"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}