{"id":247,"date":"2018-01-24T23:51:04","date_gmt":"2018-01-24T15:51:04","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=247"},"modified":"2018-01-24T23:51:04","modified_gmt":"2018-01-24T15:51:04","slug":"equilibrium-of-bodies-connected-by-cord-and-pulley-application-of-newtons-first-law-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/equilibrium-of-bodies-connected-by-cord-and-pulley-application-of-newtons-first-law-problems-and-solutions.htm","title":{"rendered":"Equilibrium of bodies connected by cord and pulley \u2013 application of Newton’s first law problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. A box of mass<\/a> 5 kg is on an inclined plane at an angle 30<\/span><\/span>o<\/span><\/span><\/sup>. The box supported by a cord. Determine the tension force (T) and the normal force<\/a> (N)!<\/span><\/span><\/p>\n

\"Equilibrium<\/p>\n

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

\"Equilibrium\u2211Fx<\/sub> = 0<\/span><\/span><\/p>\n

T \u2013 w sin 30<\/span><\/span>o<\/span><\/span><\/sup> = 0<\/span><\/span><\/p>\n

T = w sin 30<\/span><\/span>o <\/span><\/span><\/sup><\/p>\n

T = (5 kg)(9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>) sin 30<\/span><\/span>o <\/span><\/span><\/sup><\/p>\n

T = (49)(0.5) <\/span><\/span><\/p>\n

T = 24.5 Newton<\/span><\/span><\/p>\n

\u2211Fy<\/sub> = 0<\/span><\/span><\/p>\n

N \u2013 w cos 30<\/span><\/span>o<\/span><\/span><\/sup> = 0<\/span><\/span><\/p>\n

N = w cos 30<\/span><\/span>o <\/span><\/span><\/sup><\/p>\n

N = (49)(0.87) <\/span><\/span><\/p>\n

N = 43 Newton<\/span><\/span><\/p>\n

<\/p>\n

2. Two objects of mass m<\/span><\/span>1<\/span><\/sub> = m<\/span><\/span>2<\/span><\/sub> = 2 kg, connected by a massless string over a frictionless pulley. Find the tension force T<\/span><\/span>1<\/span><\/sub> and T<\/span><\/span>2<\/span><\/sub>.<\/span><\/span><\/p>\n

\"Equilibrium<\/p>\n

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

\"Equilibrium<\/p>\n

(a) Free-body diagram for object 1 (b) Free-body diagram for object 2<\/span><\/span><\/p>\n

Apply Newton’s first law to object 1 :<\/span><\/span><\/p>\n

\u2211F<\/span><\/span>y<\/span><\/sub> = 0<\/span><\/span><\/p>\n

T<\/span><\/span>1<\/span><\/sub> \u2013 w<\/span><\/span>1<\/span><\/sub> = 0<\/span><\/span><\/p>\n

T<\/span><\/span>1<\/span><\/sub> = w<\/span><\/span>1<\/span><\/sub> = m<\/span><\/span>1 <\/span><\/sub>g = (2 kg)(9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>) = 19.6 N<\/span><\/span><\/p>\n

Apply Newton’s first law<\/a> to object 2 :<\/span><\/span><\/p>\n

\u2211Fy<\/sub> = 0<\/span><\/span><\/p>\n

T<\/span><\/span>2<\/span><\/sub> \u2013 w<\/span><\/span>2<\/span><\/sub> = 0<\/span><\/span><\/p>\n

T<\/span><\/span>2<\/span><\/sub> = w<\/span><\/span>2 <\/span><\/sub>= m<\/span><\/span>2<\/span><\/span><\/sub> g = (2 kg)(9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>) = 19.6 N<\/span><\/span><\/p>\n

T<\/span><\/span>1<\/span><\/sub> = T<\/span><\/span>2 <\/span><\/sub>= 19.6 N.<\/span><\/span><\/p>\n

<\/p>\n

3. An object of weight<\/a> w<\/span><\/span>A<\/span><\/sub> = 30 N and an object of weight w<\/span><\/span>B<\/span><\/sub> = 40 N, are attached by a lightweight cord that passes over a frictionless pulley of the negligible mass. Determine the coefficient of the maximum static friction<\/a> between w<\/span><\/span>B<\/span><\/sub> and inclined surface, if the system is at rest.<\/span><\/span><\/p>\n

\"Equilibrium<\/p>\n

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

\"Equilibrium<\/p>\n

(a) Free-body diagram for object wA<\/sub> (b) Free-body diagram for object wB<\/sub><\/span><\/span><\/p>\n

Apply Newton’s first law to object w<\/span><\/span>A<\/span><\/sub> in vertical (y) direction :<\/span><\/span><\/p>\n

\u2211Fy<\/sub> = 0 (no acceleration in vertical direction)<\/span><\/span><\/p>\n

T \u2013 w<\/span><\/span>A <\/span><\/sub>= 0<\/span><\/span><\/p>\n

T = w<\/span><\/span>A<\/span><\/sub> = 30 Newton<\/span><\/span><\/p>\n

Apply Newton’s first law to object w<\/span><\/span>B <\/span><\/sub>in vertical (y) direction <\/span><\/span>:<\/span><\/span><\/p>\n

\u2211F<\/span><\/span>y<\/span><\/sub> = 0 <\/span><\/span><\/p>\n

N \u2013 w<\/span><\/span>B<\/span><\/sub> cos 45<\/span><\/span>o <\/span><\/span><\/sup>= 0<\/span><\/span><\/p>\n

N = w<\/span><\/span>B <\/span><\/sub>cos 45<\/span><\/span>o <\/span><\/span><\/sup>= (40)(0.7) = 28 Newton<\/span><\/span><\/p>\n

Apply Newton’s first law to object w<\/span><\/span>B <\/span><\/sub>in horizontal (x) direction :<\/span><\/span><\/p>\n

\u2211F<\/span><\/span>x <\/span><\/sub>= 0 <\/span><\/span><\/p>\n

F<\/span><\/span>k<\/span><\/sub> + w<\/span><\/span>B<\/span><\/sub> sin 45<\/span><\/span>o<\/span><\/span><\/sup> \u2013 T = 0<\/span><\/span><\/p>\n

\u03bc<\/span><\/span>s<\/span><\/sub> N + w<\/span><\/span>B<\/span><\/sub> sin 45<\/span><\/span>o<\/span><\/span><\/sup> \u2013 T = 0<\/span><\/span><\/p>\n

\u03bc<\/span><\/span>s<\/span><\/sub> (28) + (40)(0.7) \u2013 30 = 0<\/span><\/span><\/p>\n

\u03bc<\/span><\/span>s<\/span><\/sub> (28) + 28 \u2013 30 = 0<\/span><\/span><\/p>\n

\u03bc<\/span><\/span>s <\/span><\/sub>(28) = 30 – 28<\/span><\/span><\/p>\n

\u03bc<\/span><\/span>s <\/span><\/sub>(28) = 2<\/span><\/span><\/p>\n

\u03bc<\/span><\/span>s <\/span><\/sub>= 2 \/ 28 <\/span><\/span><\/p>\n

\u03bc<\/span><\/span>s <\/span><\/sub>= 0.07<\/span><\/span><\/p>\n

The coefficient of the maximum static friction between w<\/span><\/span>B<\/span><\/sub> and inclined surface = 0.07. <\/span><\/span><\/p>\n

[wpdm_package id=’490′]<\/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 inclined plane<\/a><\/li>\n<\/ol>\n

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

    1. A box of mass 5 kg is on an inclined plane at an angle 30o. The box supported by a cord. Determine the tension force (T) and the normal force (N)! Solution \u2211Fx = 0 T \u2013 w sin 30o = 0 T = w sin 30o T = (5 kg)(9.8 m\/s2) sin 30o … 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 bodies connected by cord and pulley \u2013 application of 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-247","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\/247","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=247"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/247\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=247"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=247"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=247"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}