{"id":828,"date":"2022-04-21T18:00:00","date_gmt":"2022-04-21T16:00:00","guid":{"rendered":"https:\/\/technischemechanik.com\/?p=828"},"modified":"2022-04-20T18:45:36","modified_gmt":"2022-04-20T16:45:36","slug":"statik-geparktes-auto-auf-abschussiger-strase-gleichgewicht","status":"publish","type":"post","link":"https:\/\/technischemechanik.com\/en\/2022\/04\/21\/statik-geparktes-auto-auf-abschussiger-strase-gleichgewicht\/","title":{"rendered":"Statik: Geparktes Auto auf absch\u00fcssiger Stra\u00dfe &#8211; Gleichgewicht"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Herzlich Willkommen!<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Wir wenden unser bisher erworbenes Wissen \u00fcber statisches Gleichgewicht heute auf ein geparktes Auto auf einer absch\u00fcssigen Stra\u00dfe an. Die Frage ist, wie gro\u00df die Bremskr\u00e4fte sein m\u00fcssen, damit das Auto auf der Stra\u00dfe stehen bleibt ohne wegzurollen.<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\"><p>Ein Sportwagen hat die Masse m und seinen Schwerpunkt in S. Die vorderen beiden Federn haben die Steifigkeit cA und die hinteren beiden cB. Bestimme die Stauchung der Federn, wenn das Auto auf einer schiefen Ebene laut Skizze geparkt wird. Welche Reibkraft FB muss auf jedes Hinterrad aufgebracht werden, um das Auto im Gleichgewicht zu halten? <\/p><p>Geg.: m = 1500kg, g = 9.81m\/s^2, cA = 58 kN\/m, cB = 65 kN\/m, a = 0.8m, b = 1.2m,c = 0.4m, \u03b1 = 30\u00b0<\/p><cite>Quelle: Aufgabe 5.32 (S. 283) aus Russell C. Hibbeler, Technische Mechanik 1 Statik, 12. Auflage, 2012 Pearson GmbH, M\u00fcnchen<\/cite><\/blockquote>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large\"><img data-recalc-dims=\"1\" loading=\"lazy\" decoding=\"async\" width=\"712\" height=\"437\" data-attachment-id=\"829\" data-permalink=\"https:\/\/technischemechanik.com\/en\/abb04-2\/\" data-orig-file=\"https:\/\/i0.wp.com\/technischemechanik.com\/wp-content\/uploads\/2022\/04\/abb04.jpg?fit=876%2C538&amp;ssl=1\" data-orig-size=\"876,538\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;,&quot;orientation&quot;:&quot;0&quot;}\" data-image-title=\"abb04\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/i0.wp.com\/technischemechanik.com\/wp-content\/uploads\/2022\/04\/abb04.jpg?fit=712%2C437&amp;ssl=1\" src=\"https:\/\/i0.wp.com\/technischemechanik.com\/wp-content\/uploads\/2022\/04\/abb04.jpg?resize=712%2C437&#038;ssl=1\" alt=\"\" class=\"wp-image-829\" \/><\/figure>\n<\/div>\n\n\n<p class=\"wp-block-paragraph\">Wir wissen nun schon, dass ein Freik\u00f6rperbild essentiell ist. Auch hier starten wir damit. Danach kann auch schon das Gleichgewicht der Kr\u00e4fte und Momente aufgestellt werden. Da wir es mit 3 unbekannten Kr\u00e4ften zu tun haben, also einem statisch bestimmten System, k\u00f6nnen wir aus den drei Gleichungen des Gleichgewichts auch alle diese Kr\u00e4fte berechnen. Zum Schluss diskutieren wir noch, wie sich Kraft und Stauchung bei einer idealen Feder im Allgemeinen verhalten und berechnen die Stauchungen der Federn an Vorder- und Hinterr\u00e4dern. Wie gewohnt findet ihr all diese Schritte im Video. <\/p>\n\n\n\n<figure class=\"wp-block-embed is-type-rich is-provider-handler-einbetten wp-block-embed-handler-einbetten wp-embed-aspect-16-9 wp-has-aspect-ratio\"><div class=\"wp-block-embed__wrapper\">\n<span class=\"embed-youtube\" style=\"text-align:center; display: block;\"><iframe loading=\"lazy\" class=\"youtube-player\" width=\"712\" height=\"401\" src=\"https:\/\/www.youtube.com\/embed\/ysiGeNwoV6w?version=3&#038;rel=1&#038;showsearch=0&#038;showinfo=1&#038;iv_load_policy=1&#038;fs=1&#038;hl=en-GB&#038;autohide=2&#038;wmode=transparent\" allowfullscreen=\"true\" style=\"border:0;\" sandbox=\"allow-scripts allow-same-origin allow-popups allow-presentation allow-popups-to-escape-sandbox\"><\/iframe><\/span>\n<\/div><figcaption><br><\/figcaption><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Bei Fragen oder Unklarheiten kommentiert bitte gerne hier oder direkt auf YouTube. \u00dcber einen Daumen hoch und ein Abo auf YouTube freue ich mich nat\u00fcrlich ebenfalls. Vielen Dank f\u00fcr eure Unterst\u00fctzung!<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Bis bald,<br>Markus <\/p>\n\n\n<div class=\"crowdsignal-feedback-wrapper\" data-crowdsignal-feedback=\"{&quot;feedbackPlaceholder&quot;:&quot;Sag mir bitte was dir besonders gefallen hat und was ich verbessern soll! 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Wir wenden unser bisher erworbenes Wissen \u00fcber statisches Gleichgewicht heute auf ein geparktes Auto auf einer absch\u00fcssigen Stra\u00dfe an. Die Frage ist, wie gro\u00df die Bremskr\u00e4fte sein m\u00fcssen, damit das Auto auf der Stra\u00dfe stehen bleibt ohne wegzurollen. Ein Sportwagen hat die Masse m und seinen Schwerpunkt in S. Die vorderen beiden Federn &hellip; <a href=\"https:\/\/technischemechanik.com\/en\/2022\/04\/21\/statik-geparktes-auto-auf-abschussiger-strase-gleichgewicht\/\" class=\"more-link\">Continue reading <span class=\"screen-reader-text\">Statik: Geparktes Auto auf absch\u00fcssiger Stra\u00dfe &#8211; Gleichgewicht<\/span><\/a><\/p>","protected":false},"author":110396568,"featured_media":657,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_coblocks_attr":"","_coblocks_dimensions":"","_coblocks_responsive_height":"","_coblocks_accordion_ie_support":"","_crdt_document":"","advanced_seo_description":"","jetpack_seo_html_title":"","jetpack_seo_noindex":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":false,"jetpack_social_options":{"image_generator_settings":{"template":"highway","default_image_id":0,"font":"","enabled":false},"version":2},"jetpack_post_was_ever_published":false},"categories":[2602546,727720248,1584145,172484],"tags":[25466,729611524,100678,414019,35890,5167337,750913306,727961753,350894,166867,113216,290680250,163250810,2517859,8207,662707,720191,1213491,143003,10126,73726871,750913302,161115,750913295,2411,1121513,727213064,729611525,2326513,2722,1268153,271082537,727961757,729611523],"class_list":["post-828","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-ubungsaufgaben","category-freikorperbild","category-gleichgewicht","category-statik","tag-beispiel","tag-ebenes-zentrales-kraftesystem","tag-einfuhrung","tag-erklart","tag-featured","tag-festigkeitslehre","tag-freikorperbild","tag-gleichgewicht","tag-grundlagen","tag-hochschule","tag-kraft","tag-kraftsystem","tag-kraftvektor","tag-krafte","tag-lernen","tag-maschinenbau","tag-mechanik","tag-nachhilfe","tag-newton","tag-physik","tag-physik-nachhilfe","tag-reduktion","tag-screencast","tag-statik","tag-studium","tag-technische-mechanik","tag-technische-mechanik-1","tag-technische-mechanik-grundlagen","tag-tm-1","tag-uni","tag-verstehen","tag-zentrales-kraftsystem","tag-zentrales-kraftesystem","tag-zentralkraftsystem"],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"https:\/\/i0.wp.com\/technischemechanik.com\/wp-content\/uploads\/2022\/02\/reduktion-wordpress.png?fit=512%2C512&ssl=1","jetpack_likes_enabled":true,"jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/p7UJuu-dm","jetpack-related-posts":[],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/technischemechanik.com\/en\/wp-json\/wp\/v2\/posts\/828","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/technischemechanik.com\/en\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/technischemechanik.com\/en\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/technischemechanik.com\/en\/wp-json\/wp\/v2\/users\/110396568"}],"replies":[{"embeddable":true,"href":"https:\/\/technischemechanik.com\/en\/wp-json\/wp\/v2\/comments?post=828"}],"version-history":[{"count":3,"href":"https:\/\/technischemechanik.com\/en\/wp-json\/wp\/v2\/posts\/828\/revisions"}],"predecessor-version":[{"id":832,"href":"https:\/\/technischemechanik.com\/en\/wp-json\/wp\/v2\/posts\/828\/revisions\/832"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/technischemechanik.com\/en\/wp-json\/wp\/v2\/media\/657"}],"wp:attachment":[{"href":"https:\/\/technischemechanik.com\/en\/wp-json\/wp\/v2\/media?parent=828"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/technischemechanik.com\/en\/wp-json\/wp\/v2\/categories?post=828"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/technischemechanik.com\/en\/wp-json\/wp\/v2\/tags?post=828"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}