{"id":259,"date":"2018-01-25T10:22:29","date_gmt":"2018-01-25T02:22:29","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=259"},"modified":"2018-01-25T10:22:29","modified_gmt":"2018-01-25T02:22:29","slug":"equilibrium-of-bodies-on-inclined-plane-application-of-newtons-first-law-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/equilibrium-of-bodies-on-inclined-plane-application-of-newtons-first-law-problems-and-solutions.htm","title":{"rendered":"Equilibrium of the bodies on a inclined plane \u2013 application of the Newton’s first law problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. A 2-kg block lies on a rough inclined plane at an angle 37<\/span><\/span>o <\/span><\/span><\/sup>to the horizontal. Determine the magnitude of the external force exerted on the block, so the block is not slides down the plane. (sin 37<\/span><\/span>o<\/span><\/span><\/sup> = 0.6, cos 37<\/span><\/span>o<\/span><\/span><\/sup> = 0.8, g = 10 m.s<\/span><\/span>-2<\/span><\/span><\/sup>, \u00b5<\/span><\/span>k<\/span><\/sub> = 0.2) <\/span><\/span><\/p>\n

\"EquilibriumKnown :<\/u><\/span><\/span><\/p>\n

Mass<\/a> (m) = 2 kg<\/span><\/span><\/p>\n

Acceleration due to gravity<\/a> (g) = 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Block’s weight<\/a> (w) = m g = (2)(10) = 20 Newton<\/span><\/span><\/p>\n

Sin 37<\/span><\/span>o<\/span><\/span><\/sup> = 0.6<\/span><\/span><\/p>\n

Cos 37<\/span><\/span>o<\/span><\/span><\/sup> = 0.8<\/span><\/span><\/p>\n

Coefficient of the kinetic friction<\/a> (\u00b5<\/span><\/span>k<\/span><\/sub>) = 0.2<\/span><\/span><\/p>\n

The y-component of the weight (w<\/span><\/span>y<\/span><\/sub>)<\/span><\/span> = w<\/span><\/span> cos 37<\/span><\/span>o <\/span><\/span><\/sup>= (20)(0.8) = 16 Newton <\/span><\/span><\/p>\n

The x-component of the weight (w<\/span><\/span>x<\/span><\/sub>)<\/span><\/span> = w sin \u03b8 = (20)(sin 37) = (20)(0.6) = 12 Newton<\/span><\/span><\/p>\n

the normal force (N) = w<\/span><\/span>y <\/span><\/sub>= 16 Newton <\/span><\/span><\/p>\n

Wanted <\/u><\/span><\/span> : The external force (F) <\/span><\/span><\/p>\n

Solution <\/u><\/span><\/span>:<\/span><\/span><\/p>\n

\"Equilibriumw<\/span><\/span>x<\/span><\/sub> = 12 Newton<\/span><\/span><\/p>\n

The force of the kinetic friction (f<\/span><\/span>k<\/span><\/sub>) <\/span><\/span>= \u00b5<\/span><\/span>k<\/span><\/sub> N = (0.1)(16) = 1.6 Newton<\/span><\/span><\/p>\n

The magnitude of the external force F exerted on the block<\/u><\/span><\/span> :<\/span><\/span><\/p>\n

F + f<\/span><\/span>k<\/span><\/sub> – w<\/span><\/span>x <\/span><\/sub>= 0 <\/span><\/span><\/p>\n

F = w<\/span><\/span>x <\/span><\/sub>– f<\/span><\/span>k<\/span><\/sub><\/p>\n

F = 12 \u2013 1.6 <\/span><\/span><\/p>\n

F = 10.4 Newton <\/span><\/span><\/p>\n

The external force F greater than 10.4 Newton.<\/span><\/span><\/p>\n

<\/p>\n

2. Mass of a block = 2 kg, coefficient of static friction \u00b5<\/span><\/span>s <\/span><\/sub>= 0.4 and \u03b8 = 45<\/span><\/span>o<\/span><\/span><\/sup>. Determine the magnitude of the force F so the block start to slides up. <\/span><\/span><\/p>\n

\"EquilibriumKnown :<\/u><\/span><\/span><\/p>\n

The coefficient of the static friction (\u00b5<\/span><\/span>s<\/span><\/sub>) = 0.4<\/span><\/span><\/p>\n

Angle (\u03b8) = 45<\/span><\/span>o<\/span><\/span><\/sup><\/p>\n

Acceleration due to gravity (g) = 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Block’s mass (m) = 2 kilogram<\/span><\/span><\/p>\n

Block’s weight (w) = m g = (2 kg)(10 m\/s<\/span><\/span>2<\/span><\/span><\/sup>) = 20 kg m\/s<\/span><\/span>2<\/span><\/span><\/sup> = 20 Newton<\/span><\/span><\/p>\n

The x-component of the weight (w<\/span><\/span>x<\/span><\/sub>)<\/span><\/span> = w sin \u03b8 = (20)(sin 45) = (20)(0.5\u221a2) = 10\u221a2 Newton<\/span><\/span><\/p>\n

The y-component of the weight (w<\/span><\/span>y<\/span><\/sub>)<\/span><\/span> = w cos \u03b8 = (20)(cos 45) = (20)(0.5\u221a2) = 10\u221a2 Newton<\/span><\/span><\/p>\n

Wanted<\/u><\/span><\/span> : The magnitude of the force F<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

\"EquilibriumBlock starts to slide up, if <\/span><\/span>F<\/i><\/span><\/span> \u2265 <\/span><\/span>w<\/i><\/span><\/span>x<\/i><\/span><\/sub> +<\/span><\/span> f<\/i><\/span><\/span>s<\/i><\/span><\/sub>. <\/span><\/span><\/p>\n

The x-component of the weight :<\/span><\/span><\/p>\n

w<\/span><\/span>x<\/span><\/sub> = 10\u221a2 Newton<\/span><\/span><\/p>\n

the y-component of the weight <\/u><\/span><\/span>:<\/span><\/span><\/p>\n

w<\/span><\/span>y<\/span><\/sub> = 10\u221a2 Newton<\/span><\/span><\/p>\n

The normal force <\/u><\/span><\/span>:<\/span><\/span><\/p>\n

N = w<\/span><\/span>y<\/span><\/sub> = 10\u221a2 Newton<\/span><\/span><\/p>\n

The force of the static friction<\/u><\/span><\/span> :<\/span><\/span><\/p>\n

f<\/i><\/span><\/span>s<\/i><\/span><\/sub> = \u00b5<\/span><\/span>s <\/span><\/sub>N = (0,4)(10\u221a2) = 4\u221a2 <\/span><\/span><\/p>\n

The magnitude of the force F so that the block starts to slide up<\/u><\/span><\/span> :<\/span><\/span><\/p>\n

F<\/i><\/span><\/span> \u2265 <\/span><\/span>w<\/i><\/span><\/span>x<\/i><\/span><\/sub> +<\/span><\/span> f<\/i><\/span><\/span>s<\/i><\/span><\/sub><\/p>\n

F<\/i><\/span><\/span> \u2265 10\u221a2 +<\/span><\/span> 4<\/i><\/span><\/span>\u221a2 <\/span><\/span><\/p>\n

F<\/i><\/span><\/span> \u2265 14\u221a2 Newton<\/span><\/span><\/p>\n

[wpdm_package id=’492′]<\/p>\n

    \n
  1. Particles in one-dimensional equilibrium<\/a><\/li>\n
  2. Particles in two-dimensional equilibrium<\/a><\/li>\n
  3. Equilibrium of bodies connected by cord and pulley<\/a><\/li>\n
  4. Equilibrium of bodies on the inclined plane<\/a><\/li>\n<\/ol>\n

    <\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

    1. A 2-kg block lies on a rough inclined plane at an angle 37o to the horizontal. Determine the magnitude of the external force exerted on the block, so the block is not slides down the plane. (sin 37o = 0.6, cos 37o = 0.8, g = 10 m.s-2, \u00b5k = 0.2) Known : Mass … Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_titles_title":"","_seopress_titles_desc":"","_seopress_robots_index":"","_seopress_robots_follow":"","_seopress_robots_imageindex":"","_seopress_robots_snippet":"","_seopress_robots_primary_cat":"","_seopress_robots_breadcrumbs":"","_seopress_robots_freeze_modified_date":"","_seopress_robots_custom_modified_date":"","_seopress_robots_canonical":"","_seopress_social_fb_title":"","_seopress_social_fb_desc":"","_seopress_social_fb_img":"","_seopress_social_fb_img_attachment_id":0,"_seopress_social_fb_img_width":0,"_seopress_social_fb_img_height":0,"_seopress_social_twitter_title":"","_seopress_social_twitter_desc":"","_seopress_social_twitter_img":"","_seopress_social_twitter_img_attachment_id":0,"_seopress_social_twitter_img_width":0,"_seopress_social_twitter_img_height":0,"_seopress_redirections_value":"","_seopress_redirections_enabled":"","_seopress_redirections_enabled_regex":"","_seopress_redirections_logged_status":"","_seopress_redirections_param":"","_seopress_redirections_type":0,"_seopress_analysis_target_kw":"Equilibrium of the bodies on a inclined plane \u2013 application of the Newton's first law problems and solutions","_seopress_news_disabled":"","_seopress_video_disabled":"","_seopress_video":[],"_seopress_pro_schemas_manual":[],"_seopress_pro_rich_snippets_disable_all":"","_seopress_pro_rich_snippets_disable":[],"_seopress_pro_schemas":[],"footnotes":""},"categories":[3],"tags":[],"class_list":["post-259","post","type-post","status-publish","format-standard","hentry","category-solved-problems-in-basic-physics"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/259","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/comments?post=259"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/259\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=259"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=259"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=259"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}