{"id":563,"date":"2020-12-01T16:09:02","date_gmt":"2020-12-01T16:09:02","guid":{"rendered":"https:\/\/pickedshares.com\/?p=563"},"modified":"2021-05-31T18:08:35","modified_gmt":"2021-05-31T18:08:35","slug":"engineering-mechanics-1-exercise-10-calculation-of-resulting-forces-and-moments-three-dimensional","status":"publish","type":"post","link":"https:\/\/pickedshares.com\/en\/engineering-mechanics-1-exercise-10-calculation-of-resulting-forces-and-moments-three-dimensional\/","title":{"rendered":"Calculation of resultant forces and moments, three-dimensional"},"content":{"rendered":"\n<h3 class=\"wp-block-heading\">Task<\/h3>\n\n\n\n<p>The forces F<sub>1<\/sub>, F<sub>2<\/sub> and F<sub>3<\/sub> act on a frame as shown. The acting forces have to be summarized in point A and point B. Calculate the resultant force and the resulting moments.<\/p>\n\n\n\n<p>given: F<sub>1<\/sub> = 2F, F<sub>2<\/sub> = F, F<sub>3<\/sub> = 2F, a<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"817\" height=\"757\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/20201104_205413.jpg\" alt=\"Frame with attacking forces, three-dimensional\" class=\"wp-image-404\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/20201104_205413.jpg 817w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/20201104_205413-300x278.jpg 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/20201104_205413-768x712.jpg 768w\" sizes=\"auto, (max-width: 817px) 100vw, 817px\" \/><figcaption>Frame with attacking forces, three-dimensional<\/figcaption><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\">Solution<\/h3>\n\n\n\n<p>The following video is in german language.<\/p>\n\n\n<figure class=\"wp-block-embed-youtube wp-block-embed is-type-video is-provider-youtube wp-embed-aspect-16-9 wp-has-aspect-ratio\"><div class=\"lyte-wrapper\" title=\"Resultierende Kraft und Moment berechnen, dreidimensional - Technische Mechanik 1, &Uuml;bung 10\" style=\"width:640px;max-width:100%;margin:5px auto;\"><div class=\"lyMe hidef\" id=\"WYL_38XOY1ehXK4\" itemprop=\"video\" itemscope itemtype=\"https:\/\/schema.org\/VideoObject\"><div><meta itemprop=\"thumbnailUrl\" content=\"https:\/\/pickedshares.com\/wp-content\/plugins\/wp-youtube-lyte\/lyteCache.php?origThumbUrl=https%3A%2F%2Fi.ytimg.com%2Fvi%2F38XOY1ehXK4%2Fmaxresdefault.jpg\" \/><meta itemprop=\"embedURL\" content=\"https:\/\/www.youtube.com\/embed\/38XOY1ehXK4\" \/><meta itemprop=\"duration\" content=\"PT13M22S\" \/><meta itemprop=\"uploadDate\" content=\"2020-11-30T21:30:48Z\" \/><\/div><meta itemprop=\"accessibilityFeature\" content=\"captions\" \/><div id=\"lyte_38XOY1ehXK4\" data-src=\"https:\/\/pickedshares.com\/wp-content\/plugins\/wp-youtube-lyte\/lyteCache.php?origThumbUrl=https%3A%2F%2Fi.ytimg.com%2Fvi%2F38XOY1ehXK4%2Fmaxresdefault.jpg\" class=\"pL\"><div class=\"tC\"><div class=\"tT\" itemprop=\"name\">Resultierende Kraft und Moment berechnen, dreidimensional - Technische Mechanik 1, \u00dcbung 10<\/div><\/div><div class=\"play\"><\/div><div class=\"ctrl\"><div class=\"Lctrl\"><\/div><div class=\"Rctrl\"><\/div><\/div><\/div><noscript><a href=\"https:\/\/youtu.be\/38XOY1ehXK4\" rel=\"nofollow\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pickedshares.com\/wp-content\/plugins\/wp-youtube-lyte\/lyteCache.php?origThumbUrl=https%3A%2F%2Fi.ytimg.com%2Fvi%2F38XOY1ehXK4%2F0.jpg\" alt=\"Resultierende Kraft und Moment berechnen, dreidimensional - Technische Mechanik 1, &Uuml;bung 10\" width=\"640\" height=\"340\" \/><br \/>Watch this video on YouTube<\/a><\/noscript><meta itemprop=\"description\" content=\"An einem Rahmen greifen die Kr\u00e4fte F1, F2 und F3 wie abgebildet an. Die angreifenden Kr\u00e4fte sind in Punkt A und Punkt B zusammenzufassen. Wie gro\u00df sind jeweils die resultierende Kraft und das resultierende Moment? Berechnung von Momentenvektoren mittels Kreuzprodukt (Sarrus-Regel). Eine \u00dcbung zur Berechnung von r\u00e4umlichen bzw. dreidimensionalen Problemen.\"><\/div><\/div><div class=\"lL\" style=\"max-width:100%;width:640px;margin:5px auto;\"><\/div><figcaption>Solution video for exercise 10<\/figcaption><\/figure>\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\newcommand{\\myvec}[1]{{\\begin{pmatrix}#1\\end{pmatrix}}} \\]\n<p>The force vectors are<\/p>\n\n<div style=\"overflow:auto;\">\n\\[\\tag{1} \\vec{F_1} = \\myvec{2F\\\\0\\\\0} , \\vec{F_2} = \\myvec{0\\\\0\\\\F} , \\vec{F_3} = \\myvec{-2F\\\\0\\\\0}\\]\n<\/div>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n\\[ \\newcommand{\\myvec}[1]{{\\begin{pmatrix}#1\\end{pmatrix}}} \\]\n<p>The position vectors have to be defined separately for the two cases to be considered. For the resulting moment at point A they are<\/p>\n\n<div style=\"overflow:auto;\">\n\\[\\tag{2} \\vec{r_{1A}} = \\myvec{0\\\\0\\\\4a} , \\vec{r_{2A}} = \\myvec{3a\\\\0\\\\4a} , \\vec{r_{3A}} = \\myvec{3a\\\\3a\\\\4a}\\]\n<\/div>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n\\[ \\newcommand{\\myvec}[1]{{\\begin{pmatrix}#1\\end{pmatrix}}} \\]\n<p>For the resulting moment in point B they are<\/p>\n\n<div style=\"overflow:auto;\">\n\\[\\tag{3} \\vec{r_{1B}} = \\myvec{0\\\\0\\\\0} , \\vec{r_{2B}} = \\myvec{3a\\\\0\\\\0} , \\vec{r_{3B}} = \\myvec{3a\\\\3a\\\\0}\\]\n<\/div>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n\\[ \\newcommand{\\myvec}[1]{{\\begin{pmatrix}#1\\end{pmatrix}}} \\]\n<p>The resulting force is the same regardless of whether it is calculated for A or B:<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[\\tag{4} \\sum \\vec{F_A} = \\sum \\vec{F_B} = \\vec{F_1} + \\vec{F_2} + \\vec{F_3} \\]<\/p>\n<p>\\[\\tag{5} \\sum \\vec{F} = \\myvec{2F\\\\0\\\\0} + \\myvec{0\\\\0\\\\F} + \\myvec{-2F\\\\0\\\\0} \\]<\/p>\n<p>\\[\\tag{6} \\sum \\vec{F} = \\myvec{0\\\\0\\\\F} \\]<\/p>\n<\/div>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n\\[ \\newcommand{\\myvec}[1]{{\\begin{pmatrix}#1\\end{pmatrix}}} \\]\n<p>The resulting moment at point A results from the sum of the cross products of the force and position vector, based on A. The procedure for calculating the cross product is described in the video above. The Sarrus rule is applied there.<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[\\tag{7} \\sum \\vec{M_A} = \\vec{r_{1A}} \\times \\vec{F_1} +  \\vec{r_{2A}} \\times \\vec{F_2}  + \\vec{r_{3A}} \\times \\vec{F_3}  \\]<\/p>\n<p>\\[\\tag{8} \\sum \\vec{M_A} = \\myvec{0\\\\0\\\\4a} \\times \\myvec{2F\\\\0\\\\0} + \\myvec{3a\\\\0\\\\4a} \\times \\myvec{0\\\\0\\\\F}  + \\myvec{3a\\\\3a\\\\4a} \n\\times \\myvec{-2F\\\\0\\\\0}  \\]<\/p>\n<p>\\[\\tag{9} \\sum \\vec{M_A} = \\myvec{0\\\\-3Fa\\\\6Fa} \\]<\/p>\n<\/div>\n\n\n\n<script src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-AMS_HTML\" async=\"async\">  \/\/ A comment that hinders wxWidgets from optimizing this tag too much.\n<\/script>\n\\[ \\require{cancel} \\]\n\\[ \\newcommand{\\myvec}[1]{{\\begin{pmatrix}#1\\end{pmatrix}}} \\]\n<p>For the resulting moment at point B, the equation is<\/p>\n\n<div style=\"overflow:auto;\">\n<p>\\[\\tag{10} \\sum \\vec{M_B} = \\vec{r_{1B}} \\times \\vec{F_1} +  \\vec{r_{2B}} \\times \\vec{F_2}  + \\vec{r_{3B}} \\times \\vec{F_3} \\]<\/p>\n<p>\\[\\tag{11} \\sum \\vec{M_B} = \\myvec{0\\\\0\\\\0} \\times \\myvec{2F\\\\0\\\\0}   + \\myvec{3a\\\\0\\\\0} \\times \\myvec{0\\\\0\\\\F} + \\myvec{3a\\\\3a\\\\0} \\times \\myvec{-2F\\\\0\\\\0} \\]<\/p>\n<p>\\[\\tag{12} \\sum \\vec{M_B} = \\myvec{0\\\\-3Fa\\\\6Fa} \\]<\/p>\n<\/div>\n\n\n\n<p>The fact that these two moments are equal in this case is due to the geometric arrangement in this exercise, i.e. it is more likely that the consideration of different points also leads to different moment vectors.<\/p>\n","protected":false},"excerpt":{"rendered":"<p> ... <a title=\"Calculation of resultant forces and moments, three-dimensional\" class=\"read-more\" href=\"https:\/\/pickedshares.com\/en\/engineering-mechanics-1-exercise-10-calculation-of-resulting-forces-and-moments-three-dimensional\/\" aria-label=\"Read more about Calculation of resultant forces and moments, three-dimensional\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":405,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[21,60],"tags":[26,47,48,29,31],"class_list":["post-563","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-engineering-mechanics-i","category-exercises","tag-cross-product","tag-resultant-force","tag-resulting-moments","tag-statics","tag-vector-calculation","infinite-scroll-item","masonry-post","generate-columns","tablet-grid-50","mobile-grid-100","grid-parent","grid-33"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Calculation of resultant forces and moments, three-dimensional &#8226; pickedshares<\/title>\n<meta name=\"description\" content=\"Several forces act at a three-dimensional frame. 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The resulting moment for two different points has to be calculated. 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