{"id":765,"date":"2020-12-08T15:34:48","date_gmt":"2020-12-08T15:34:48","guid":{"rendered":"https:\/\/pickedshares.com\/?p=765"},"modified":"2021-05-08T12:05:40","modified_gmt":"2021-05-08T12:05:40","slug":"engineering-mechanics-1-exercise-23-determine-the-balanced-state","status":"publish","type":"post","link":"https:\/\/pickedshares.com\/en\/engineering-mechanics-1-exercise-23-determine-the-balanced-state\/","title":{"rendered":"Determine the balanced state"},"content":{"rendered":"\n<p>This exercise is about the following questions:<\/p>\n\n\n\n<ul class=\"wp-block-list\" id=\"block-78ff143f-920b-4d07-8061-082b660ff5a6\"><li>Which forces occur at frictionless contact points?<\/li><li>How do you establish equilibrium conditions?<\/li><li>How do you calculate lever arms for forces on inclined objects?<\/li><\/ul>\n\n\n\n<h2 class=\"wp-block-heading\">Task<\/h2>\n\n\n\n<p>An ideally smooth rod with its own weight G should be inserted into a slot with the width B, so that the contact shown in A and B results and the state of equilibrium is established. What length l must the rod have so that it is in equilibrium at the angle Alpha?<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"832\" height=\"909\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201114-190352_Sketch-e1605518541320.jpg\" alt=\"Frictionless stick in balance\" class=\"wp-image-453\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201114-190352_Sketch-e1605518541320.jpg 832w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201114-190352_Sketch-e1605518541320-275x300.jpg 275w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201114-190352_Sketch-e1605518541320-768x839.jpg 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201114-190352_Sketch-e1605518541320-400x437.jpg 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201114-190352_Sketch-e1605518541320-800x874.jpg 800w\" sizes=\"auto, (max-width: 832px) 100vw, 832px\" \/><figcaption>Frictionless stick in balance<\/figcaption><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\">Solution<\/h2>\n\n\n\n<p>The following video is in German language, but English subtitles are available.<\/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=\"Gleichgewichtslage eines (ideal glatten) Stabs ermitteln - Technische Mechanik 1, &Uuml;bung 23\" style=\"width:640px;max-width:100%;margin:5px auto;\"><div class=\"lyMe hidef\" id=\"WYL_Zr7-nnPadPs\" 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%2FZr7-nnPadPs%2Fmaxresdefault.jpg\" \/><meta itemprop=\"embedURL\" content=\"https:\/\/www.youtube.com\/embed\/Zr7-nnPadPs\" \/><meta itemprop=\"duration\" content=\"PT6M31S\" \/><meta itemprop=\"uploadDate\" content=\"2021-01-01T17:16:26Z\" \/><\/div><meta itemprop=\"accessibilityFeature\" content=\"captions\" \/><div id=\"lyte_Zr7-nnPadPs\" data-src=\"https:\/\/pickedshares.com\/wp-content\/plugins\/wp-youtube-lyte\/lyteCache.php?origThumbUrl=https%3A%2F%2Fi.ytimg.com%2Fvi%2FZr7-nnPadPs%2Fmaxresdefault.jpg\" class=\"pL\"><div class=\"tC\"><div class=\"tT\" itemprop=\"name\">Gleichgewichtslage eines (ideal glatten) Stabs ermitteln - Technische Mechanik 1, \u00dcbung 23<\/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\/Zr7-nnPadPs\" 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%2FZr7-nnPadPs%2F0.jpg\" alt=\"Gleichgewichtslage eines (ideal glatten) Stabs ermitteln - Technische Mechanik 1, &Uuml;bung 23\" width=\"640\" height=\"340\" \/><br \/>Watch this video on YouTube<\/a><\/noscript><meta itemprop=\"description\" content=\"Ein ideal glatter Stab mit der L\u00e4nge l soll in einen Schlitz der Breite b gestellt werden. Zu ermitteln ist die L\u00e4nge l, die der Stab ben\u00f6tigt, um unter dem gegebenen Winkel Alpha im Schlitz stehen zu bleiben. Im Video werden die Reaktionskr\u00e4fte des Stabes angetragen (also der Stab freigeschnitten) und anschlie\u00dfend Kr\u00e4ftegleichgewicht und Momentengleichgewicht aufgestellt. Daraus kann anschlie\u00dfend die gesuchte L\u00e4nge des Stabs berechnet werden. Etwas tricky ist es, die Hebelarme f\u00fcr die Kr\u00e4fte richtig zu bestimmen, da man dabei schnell mit den Winkelfunktionen durcheinanderkommt. Leider funktioniert mein Greenscreen, der im Moment noch ein Bluescreen ist, noch nicht so richtig, d.h. es gibt manchmal etwas krisselige Stellen um meinen Kopf herum. Ich hoffe, das Video ist trotzdem interessant. Alle Aufgaben, die hier auf Youtube gepostet werden, sind auch auf der Webseite pickedshares.com nachzulesen. Der auf der Webseite beschriebene L\u00f6sungsweg f\u00fcr diese Aufgabe ist etwas anders und auch etwas schwieriger, als der hier im Video gezeigte Weg.\"><\/div><\/div><div class=\"lL\" style=\"max-width:100%;width:640px;margin:5px auto;\"><\/div><figcaption>Determination of the balanced state<\/figcaption><\/figure>\n\n\n<p>The way the problem is solved in the video is a little bit easier than the steps shown below. <\/p>\n\n\n\n<p>The property \"frictionless\" results in the following reaction forces in the points A and B.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"866\" height=\"893\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201116-094503_Sketch.jpg\" alt=\"Reaction forces in the slot\" class=\"wp-image-457\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201116-094503_Sketch.jpg 866w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201116-094503_Sketch-291x300.jpg 291w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201116-094503_Sketch-768x792.jpg 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201116-094503_Sketch-400x412.jpg 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Screenshot_20201116-094503_Sketch-800x825.jpg 800w\" sizes=\"auto, (max-width: 866px) 100vw, 866px\" \/><figcaption>Reaction forces in the slot<\/figcaption><\/figure>\n\n\n\n<!DOCTYPE html>\n<html lang=\"\">\n  <head>\n    \n    \n    <meta http-equiv=\"Content-Type\" content=\"text\/html; charset=utf-8\"\/>\n    <script type=\"text\/x-mathjax-config\">  MathJax.Hub.Config({\n    displayAlign: \"left\",\n    context: \"MathJax\",\n    TeX: {TagSide: \"left\"}\n  })\n<\/script>\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    \n  <\/head>\n  <body>\n    <!-- ****************************************************** -->\n    <!-- *        Created with wxMaxima version 20.06.6       * -->\n    <!-- ****************************************************** -->\n    <noscript>\n      <div class=\"error message\">\n        <p>Please enable JavaScript in order to get a 2d display of the equations embedded in this web page.<\/p>\n      <\/div>\n    <\/noscript>\n    <p hidden=\"hidden\">\\(      \\DeclareMathOperator{\\abs}{abs}\n      \\newcommand{\\ensuremath}[1]{\\mbox{$#1$}}\n\\)<\/p>\n    \n    <!-- Text cell -->\n    <div class=\"comment\">Balance of forces in x-direction:\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{1} 0={F_A} \\sin{\\left( \\alpha \\right) }-{F_B}\\]\n<\/p>\n    <!-- Text cell -->\n    <div class=\"comment\">Balance of forces in y-direction:\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{2} 0={F_A} \\cos{\\left( \\alpha \\right) }-G\\]\n<\/p>\n    <!-- Text cell -->\n    <div class=\"comment\">Balance of moments around A\n<\/div>\n\n<div style=\"overflow:auto;\">\n    <!-- Code cell -->\n    <p>\\[\\tag{3} 0=G\\, \\left( \\frac{l \\cos{\\left( \\alpha \\right) }}{2}-b\\right) -\\frac{{F_B} b \\sin{\\left( \\alpha \\right) }}{\\cos{\\left( \\alpha \\right) }}\\]\n<\/p>\n<\/div>\n    <!-- Text cell -->\n    <div class=\"comment\">Solve for l\n<\/div>\n\n<div style=\"overflow:auto;\">\n    <!-- Code cell -->\n    <p>\\[\\tag{4} l=\\frac{2 {F_B} b \\sin{\\left( \\alpha \\right) }+2 G b \\cos{\\left( \\alpha \\right) }}{G\\, {{\\cos{\\left( \\alpha \\right) }}^{2}}}\\]\n<\/p>\n<\/div>\n    <!-- Text cell -->\n    <div class=\"comment\">Solving equations 1 and 2 for F<sub>A<\/sub> and F<sub>B<\/sub>\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{5} {F_B}={F_A} \\sin{\\left( \\alpha \\right) }\\]\n<\/p>\n    <!-- Code cell -->\n    <p>\\[\\tag{6} {F_A}=\\frac{G}{\\cos{\\left( \\alpha \\right) }}\\]\n<\/p>\n    <!-- Text cell -->\n    <div class=\"comment\">Substitution of F<sub>A<\/sub>\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{7} {F_B}=\\frac{G \\sin{\\left( \\alpha \\right) }}{\\cos{\\left( \\alpha \\right) }}\\]\n<\/p>\n    <!-- Text cell -->\n    <div class=\"comment\">Substitution of F<sub>B<\/sub> in equation 4\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{8} l=\\frac{\\frac{2 G b\\, {{\\sin{\\left( \\alpha \\right) }}^{2}}}{\\cos{\\left( \\alpha \\right) }}+2 G b \\cos{\\left( \\alpha \\right) }}{G\\, {{\\cos{\\left( \\alpha \\right) }}^{2}}}\\]\n<\/p>\n    <!-- Text cell -->\n    <div class=\"comment\">Simplified:\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{9} l=\\frac{2 b\\, {{\\sin{\\left( \\alpha \\right) }}^{2}}+2 b\\, {{\\cos{\\left( \\alpha \\right) }}^{2}}}{{{\\cos{\\left( \\alpha \\right) }}^{3}}}\\]\n<\/p>\n    <!-- Text cell -->\n    <div class=\"comment\">Alternative notation of the result\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{10} l=\\frac{2 b\\, \\left( {{\\tan{\\left( \\alpha \\right) }}^{2}}+1\\right) }{\\cos{\\left( \\alpha \\right) }}\\]\n<\/p>\n    \n    \n  <\/body>\n<\/html>\n","protected":false},"excerpt":{"rendered":"<p> ... <a title=\"Determine the balanced state\" class=\"read-more\" href=\"https:\/\/pickedshares.com\/en\/engineering-mechanics-1-exercise-23-determine-the-balanced-state\/\" aria-label=\"Read more about Determine the balanced state\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":451,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[21,60],"tags":[45,46,29],"class_list":["post-765","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-engineering-mechanics-i","category-exercises","tag-bearing-loads","tag-reaction-forces","tag-statics","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>Determine the balanced state &#8226; 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