{"id":224,"date":"2018-01-23T18:00:03","date_gmt":"2018-01-23T10:00:03","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=224"},"modified":"2018-01-23T18:00:03","modified_gmt":"2018-01-23T10:00:03","slug":"bodies-connected-by-cord-and-pulley-application-of-newtons-law-of-motion-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/bodies-connected-by-cord-and-pulley-application-of-newtons-law-of-motion-problems-and-solutions.htm","title":{"rendered":"Bodies connected by the cord and pulley – application of Newton’s law of motion problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. Two boxes are connected by a cord running over a pulley. Ignore the mass of the cord and pulley and any friction in the pulley. Mass<\/a> of the box 1 = 2 kg, mass of the box 2 = 3 kg, acceleration due to gravity<\/a> = 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup>. Find <\/span><\/span>(a) The acceleration of the system (b) The tension in the cord! <\/span><\/span><\/p>\n

\"Bodies<\/p>\n

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

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

Mass of the box 1 (m<\/span><\/span>1<\/span><\/sub>) = 2 kg<\/span><\/span><\/p>\n

Mass of the box 2 (m<\/span><\/span>2<\/span><\/sub>) = 3 kg<\/span><\/span><\/p>\n

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

Weight<\/a> of the box 1 (w<\/span><\/span>1<\/span><\/sub>) = m<\/span><\/span>1<\/span><\/sub> g = (2)(10) = 20 Newton<\/span><\/span><\/p>\n

Weight of the box 2 (w<\/span><\/span>2<\/span><\/sub>) = m<\/span><\/span>2<\/span><\/sub> g = (3)(10) = 30 Newton<\/span><\/span><\/p>\n

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

(a) magnitude and direction of the acceleration <\/u><\/span><\/span><\/p>\n

w<\/span><\/span>2 <\/span><\/sub>> w<\/span><\/span>1<\/span><\/sub> so the <\/span><\/span>box 2 accelerates downward and the box 1 accelerates upward. <\/span><\/span><\/p>\n

Forces that has the same direction with acceleration (w<\/span><\/span>2<\/span><\/sub> and T<\/span><\/span>1<\/span><\/sub>), its sign positive. Forces that has opposite direction with acceleration (T<\/span><\/span>2<\/span><\/sub> and w<\/span><\/span>1<\/span><\/sub>), its sign negative.<\/span><\/span><\/p>\n

\u2211F = m a<\/span><\/span><\/p>\n

w<\/span><\/span>2<\/span><\/sub> \u2013 T<\/span><\/span>2<\/span><\/sub> + T<\/span><\/span>1<\/span><\/span><\/sub> – w<\/span><\/span>1<\/span><\/sub> = (m<\/span><\/span>1<\/span><\/sub> + m<\/span><\/span>2<\/span><\/sub>) a ——-> T<\/span><\/span>1<\/span><\/sub> = T<\/span><\/span>2<\/span><\/sub> = T<\/span><\/span><\/p>\n

w<\/span><\/span>2<\/span><\/sub> \u2013 T + T – w<\/span><\/span>1<\/span><\/sub> = (m<\/span><\/span>1<\/span><\/sub> + m<\/span><\/span>2<\/span><\/sub>) a<\/span><\/span><\/p>\n

w<\/span><\/span>2<\/span><\/sub> \u2013 w<\/span><\/span>1<\/span><\/sub> = (m<\/span><\/span>1<\/span><\/sub> + m<\/span><\/span>2<\/span><\/sub>) a<\/span><\/span><\/p>\n

30 \u2013 20 = (2 + 3) a<\/span><\/span><\/p>\n

10 = 5 a<\/span><\/span><\/p>\n

a = 10 \/ 5<\/span><\/span><\/p>\n

a = 2 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Magnitude of the acceleration<\/a> is 2 m\/s<\/span><\/span>2<\/span><\/span><\/sup>.<\/span><\/span><\/p>\n

(b) The tension force<\/u><\/span><\/span><\/p>\n

The box 2 :<\/u><\/span><\/span><\/p>\n

There are two forces acts on the box 2 : first, weight of the box 2 (w<\/span><\/span>2<\/span><\/sub>), points downward so it’s positive. Second, tension force exerted on the box 2 (T<\/span><\/span>2<\/span><\/sub>), points upward so it’s negative. Apply Newton’s second law<\/a> of motion.<\/span><\/span><\/p>\n

\u2211F = m a<\/span><\/span><\/p>\n

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

30 \u2013 T<\/span><\/span>2<\/span><\/sub> = (3)(2)<\/span><\/span><\/p>\n

30 \u2013 T<\/span><\/span>2<\/span><\/sub> = 6<\/span><\/span><\/p>\n

T<\/span><\/span>2<\/span><\/sub> = 30 \u2013 6<\/span><\/span><\/p>\n

T<\/span><\/span>2<\/span><\/sub> = 24 Newton<\/span><\/span><\/p>\n

Box 1 :<\/u><\/span><\/span><\/p>\n

There are two forces acts on the box 1. <\/span><\/span>First<\/i><\/span><\/span>, weight of the box 1 (w<\/span><\/span>1<\/span><\/sub>), points downward so it’s negative. <\/span><\/span>Second<\/i><\/span><\/span>, the tension force exerted on the box 1<\/span><\/span> (T<\/span><\/span>1<\/span><\/sub>) points upward so it’s positive. Apply Newton’s second law of motion : <\/span><\/span><\/p>\n

\u2211F = m a<\/span><\/span><\/p>\n

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

T<\/span><\/span>1<\/span><\/sub> \u2013 20 = (2)(2)<\/span><\/span><\/p>\n

T<\/span><\/span>1<\/span><\/sub> \u2013 20 = 4<\/span><\/span><\/p>\n

T<\/span><\/span>1<\/span><\/sub> = 20 + 4<\/span><\/span><\/p>\n

T<\/span><\/span>1<\/span><\/sub> = 24 Newton<\/span><\/span><\/p>\n

Magnitude of the tension force = T<\/span><\/span>1<\/span><\/sub> = T<\/span><\/span>2<\/span><\/sub> = T = 24 Newton<\/span><\/span><\/p>\n

<\/p>\n

2. An object on a rough horizontal surface. Mass of the object 1 = 2 kg, mass of the object 2 = 4 kg, acceleration due to gravity = 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup>, coefficient of the static friction = 0.4, coefficient of the kinetic friction = 0.3. The system is at rest or accelerated ? If the system is accelerated, find the magnitude and direction of the system’s acceleration!<\/span><\/span><\/p>\n

\"Bodies<\/p>\n

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

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

Mass of the object 1 (m<\/span><\/span>1<\/span><\/sub>) = 2 kg<\/span><\/span><\/p>\n

Mass of the object 2 (m<\/span><\/span>2<\/span><\/sub>) = 4 kg<\/span><\/span><\/p>\n

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

Coefficient of the static friction<\/a><\/span><\/span> (<\/span><\/span>\u03bc<\/span><\/span>s<\/span><\/sub>) = 0.4<\/span><\/span><\/p>\n

The coefficient of the kinetic friction <\/span><\/span>(<\/span><\/span>\u03bc<\/span><\/span>k<\/span><\/sub>)<\/span><\/span> = 0.3<\/span><\/span><\/p>\n

Weight of the object 1 (w<\/span><\/span>1<\/span><\/sub>) = m<\/span><\/span>1<\/span><\/sub> g = (2)(10) = 20 Newton<\/span><\/span><\/p>\n

Weight of the object 2 (w<\/span><\/span>2<\/span><\/sub>) = m<\/span><\/span>2<\/span><\/sub> g = (4)(10) = 40 Newton<\/span><\/span><\/p>\n

Normal force<\/a> exerted on the object 1 (N) = w<\/span><\/span>1<\/span><\/sub> = 20 Newton<\/span><\/span><\/p>\n

Force of the static friction exerted on the object 1 (f<\/span><\/span>s<\/span><\/span><\/sub>) = <\/span><\/span>\u03bc<\/span><\/span>s <\/span><\/sub>N = (0.4)(20) = 8 Newton<\/span><\/span><\/p>\n

Force of the kinetic friction exerted on the object 1 (f<\/span><\/span>k<\/span><\/sub>) = <\/span><\/span>\u03bc<\/span><\/span>k <\/span><\/sub>N = (0.3)(20) = 6 Newton<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> acceleration (a)<\/span><\/span><\/p>\n

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

w<\/span><\/span>2<\/span><\/sub> > f<\/span><\/span>s<\/span><\/sub> (40 Newton > 8 Newton) so the object 2 is accelerated vertically downward and the object 1 is accelerated horizontally rightward. The friction force that acts on the objects 1 is the force of the kinetic friction (f<\/span><\/span>k<\/span><\/sub>). Apply Newton’s second law of motion :<\/span><\/span><\/p>\n

\u2211F = m a<\/span><\/span><\/p>\n

w<\/span><\/span>2<\/span><\/sub> \u2013 the = (m<\/span><\/span>1<\/span><\/sub> + m<\/span><\/span>2<\/span><\/sub>) a<\/span><\/span><\/p>\n

40 \u2013 6 = (2 + 4) a<\/span><\/span><\/p>\n

34 = 6 a<\/span><\/span><\/p>\n

a = 34 \/ 6 = 17 \/ 3<\/span><\/span><\/p>\n

a = 5.7 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Magnitude of the acceleration = 5.7 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

[wpdm_package id=’484′]<\/p>\n

    \n
  1. Mass and weight<\/a><\/li>\n
  2. Normal force<\/a><\/li>\n
  3. Newton’s second law of motion<\/a><\/li>\n
  4. Friction force<\/a><\/li>\n
  5. Motion on horizontal surface without friction force<\/a><\/li>\n
  6. The motion of two bodies with the same acceleration on the rough horizontal surface with the friction force<\/a><\/li>\n
  7. Motion on the inclined plane without friction force<\/a><\/li>\n
  8. Motion on the rough inclined plane with the friction force<\/a><\/li>\n
  9. Motion in an elevator<\/a><\/li>\n
  10. The motion of bodies connected by cord and pulley<\/a><\/li>\n
  11. Two bodies with the same magnitude of accelerations<\/a><\/li>\n
  12. Rounding a flat curve – dynamics of circular motion<\/a><\/li>\n
  13. Rounding a banked curve – dynamics of circular motion<\/a><\/li>\n
  14. Uniform motion in a horizontal circle<\/a><\/li>\n
  15. Centripetal force in uniform circular motion<\/a><\/li>\n<\/ol>\n

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

    1. Two boxes are connected by a cord running over a pulley. Ignore the mass of the cord and pulley and any friction in the pulley. Mass of the box 1 = 2 kg, mass of the box 2 = 3 kg, acceleration due to gravity = 10 m\/s2. Find (a) The acceleration of the … 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":"Bodies connected by the cord and pulley - application of Newton's law of motion 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-224","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\/224","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=224"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/224\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=224"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=224"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=224"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}