{"id":768,"date":"2020-12-08T16:19:33","date_gmt":"2020-12-08T16:19:33","guid":{"rendered":"https:\/\/pickedshares.com\/?p=768"},"modified":"2021-05-08T12:07:57","modified_gmt":"2021-05-08T12:07:57","slug":"engineering-mechanics-1-exercise-25-car-on-a-sloping-road","status":"publish","type":"post","link":"https:\/\/pickedshares.com\/en\/engineering-mechanics-1-exercise-25-car-on-a-sloping-road\/","title":{"rendered":"Car on a sloping road"},"content":{"rendered":"\n<h2 class=\"wp-block-heading\">Task<\/h2>\n\n\n\n<p>A car is parked on a sloping road with the handbrake on. The handbrake acts on the rear wheels. What is the minimum <a href=\"https:\/\/pickedshares.com\/en\/static-and-sliding-friction-values\/\">coefficient of static friction<\/a> required to prevent the car from slipping?<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"722\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/TM1-25-1024x722.jpg\" alt=\"Car on a sloping road\" class=\"wp-image-459\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/TM1-25-1024x722.jpg 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/TM1-25-300x212.jpg 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/TM1-25-768x541.jpg 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/TM1-25-400x282.jpg 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/TM1-25-800x564.jpg 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/TM1-25.jpg 1329w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Car on a sloping road<\/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-4-3 wp-has-aspect-ratio\"><div class=\"lyte-wrapper\" title=\"Technische Mechanik 1, &Uuml;bung 25 - erforderlichen Haftreibungskoeffizienten berechnen\" style=\"width:640px;max-width:100%;margin:5px auto;\"><div class=\"lyMe hidef\" id=\"WYL_3TaLDeBc35g\" 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%2F3TaLDeBc35g%2Fmaxresdefault.jpg\" \/><meta itemprop=\"embedURL\" content=\"https:\/\/www.youtube.com\/embed\/3TaLDeBc35g\" \/><meta itemprop=\"duration\" content=\"PT5M36S\" \/><meta itemprop=\"uploadDate\" content=\"2021-01-16T17:00:04Z\" \/><\/div><meta itemprop=\"accessibilityFeature\" content=\"captions\" \/><div id=\"lyte_3TaLDeBc35g\" data-src=\"https:\/\/pickedshares.com\/wp-content\/plugins\/wp-youtube-lyte\/lyteCache.php?origThumbUrl=https%3A%2F%2Fi.ytimg.com%2Fvi%2F3TaLDeBc35g%2Fmaxresdefault.jpg\" class=\"pL\"><div class=\"tC\"><div class=\"tT\" itemprop=\"name\">Technische Mechanik 1, \u00dcbung 25 - erforderlichen Haftreibungskoeffizienten berechnen<\/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\/3TaLDeBc35g\" 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%2F3TaLDeBc35g%2F0.jpg\" alt=\"Technische Mechanik 1, &Uuml;bung 25 - erforderlichen Haftreibungskoeffizienten berechnen\" width=\"640\" height=\"340\" \/><br \/>Watch this video on YouTube<\/a><\/noscript><meta itemprop=\"description\" content=\"Wie gro\u00df muss der Haftreibungskoeffizient mindestens sein, damit das auf einer absch\u00fcssigen Stra\u00dfe geparkte Auto nicht ins Rutschen kommt? Die Handbremse ist angezogen und wirkt auf die Hinterr\u00e4der. Eine \u00dcbung aus der technischen Mechanik 1 zum Thema Reibung. Es werden die statischen Gleichgewichtsbedingungen aufgestellt und um den Zusammenhang zwischen Reibkraft und Normalkraft erg\u00e4nzt. Eine &quot;Besonderheit&quot; ist, dass das Koordinatensystem um den Neigungswinkel gedreht ist, damit die Berechnung der Kr\u00e4ftebilanzen einfacher ausf\u00e4llt. Die \u00dcbung gibt es hier zum Nachlesen: https:\/\/pickedshares.com\/technische-mechanik-1-uebung-25-auto-auf-abschuessiger-strasse\/\"><\/div><\/div><div class=\"lL\" style=\"max-width:100%;width:640px;margin:5px auto;\"><\/div><figcaption>The solution video with German audio, but English subtitles<\/figcaption><\/figure>\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"607\" src=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Tm1-25solved-1024x607.jpg\" alt=\"Reaction forces on the wheels\" class=\"wp-image-461\" srcset=\"https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Tm1-25solved-1024x607.jpg 1024w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Tm1-25solved-300x178.jpg 300w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Tm1-25solved-768x456.jpg 768w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Tm1-25solved-400x237.jpg 400w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Tm1-25solved-800x475.jpg 800w, https:\/\/pickedshares.com\/wp-content\/uploads\/2020\/11\/Tm1-25solved.jpg 1492w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><figcaption>Reaction forces on the wheels<\/figcaption><\/figure>\n\n\n\n<p>The coordinate system will be turned by the angle Alpha for the solution.<\/p>\n\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    <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    <!-- Text cell -->\n    <div class=\"comment\">Balance of forces in x-direction\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{1} 0={F_{\\mathit{Bx}}}-G \\sin{\\left( \\alpha \\right) }\\]\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=-G \\cos{\\left( \\alpha \\right) }+{F_{\\mathit{By}}}+{F_A}\\]\n<\/p>\n    <!-- Text cell -->\n    <div class=\"comment\">Balance of forces around A (left-turning moments are positive)\n<\/div>\n    <!-- Code cell -->\n\n<div style=\"overflow:auto;\">\n    <p>\\[\\tag{3} 0=G a \\sin{\\left( \\alpha \\right) }-G b \\cos{\\left( \\alpha \\right) }+{F_{\\mathit{By}}} \\left( c+b\\right) \\]\n<\/p>\n<\/div>\n    <!-- Text cell -->\n    <div class=\"comment\">Relationship between tangential and normal force on the rear wheel\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{4} {F_{\\mathit{Bx}}}={F_{\\mathit{By}}} {\u00b5_0}\\]\n<\/p>\n    <!-- Text cell -->\n    <div class=\"comment\">Solved for F<sub>By<\/sub> and used in the balance of moments\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{5} {F_{\\mathit{By}}}=\\frac{{F_{\\mathit{Bx}}}}{{\u00b5_0}}\\]\n<\/p>\n<div style=\"overflow:auto;\">\n    <!-- Code cell -->\n    <p>\\[\\tag{6} 0=G a \\sin{\\left( \\alpha \\right) }-G b \\cos{\\left( \\alpha \\right) }+\\frac{{F_{\\mathit{Bx}}} \\left( c+b\\right) }{{\u00b5_0}}\\]\n<\/p><\/div>\n    <!-- Text cell -->\n    <div class=\"comment\">Equation 1 solved for F<sub>Bx<\/sub> and used in the previous result\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{7} {F_{\\mathit{Bx}}}=G \\sin{\\left( \\alpha \\right) }\\]\n<\/p>\n    <!-- Code cell -->\n\n<div style=\"overflow:auto;\">\n    <p>\\[\\tag{8} 0=\\frac{G\\, \\left( c+b\\right)  \\sin{\\left( \\alpha \\right) }}{{\u00b5_0}}+G a \\sin{\\left( \\alpha \\right) }-G b \\cos{\\left( \\alpha \\right) }\\]\n<\/p>\n<\/div>\n    <!-- Text cell -->\n    <div class=\"comment\">The minimum value for the coefficient of static friction is\n<\/div>\n    <!-- Code cell -->\n    <p>\\[\\tag{9} {\u00b5_0}=\\frac{\\left( c+b\\right)  \\sin{\\left( \\alpha \\right) }}{b \\cos{\\left( \\alpha \\right) -}a \\sin{\\left( \\alpha \\right) }}\\]\n<\/p>\n\n\n\n<p>It is obvious that the weight does not play a role in the determination of the required coefficient of static friction, i.e. a discussion of the distribution of the weight force over several wheels per axle is superfluous.<\/p>\n","protected":false},"excerpt":{"rendered":"<p> ... <a title=\"Car on a sloping road\" class=\"read-more\" href=\"https:\/\/pickedshares.com\/en\/engineering-mechanics-1-exercise-25-car-on-a-sloping-road\/\" aria-label=\"Read more about Car on a sloping road\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":460,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[21,60],"tags":[45,27,46,29],"class_list":["post-768","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-engineering-mechanics-i","category-exercises","tag-bearing-loads","tag-friction","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>Car on a sloping road &#8226; 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