{"id":195,"date":"2018-01-19T07:49:00","date_gmt":"2018-01-18T23:49:00","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=195"},"modified":"2018-01-19T07:49:00","modified_gmt":"2018-01-18T23:49:00","slug":"motion-of-two-bodies-with-the-same-accelerations-on-rough-horizontal-surface-with-friction-force-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/motion-of-two-bodies-with-the-same-accelerations-on-rough-horizontal-surface-with-friction-force-problems-and-solutions.htm","title":{"rendered":"Motion of two bodies with the same accelerations on the rough horizontal surface with the friction force – problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. Mass<\/a> of the box 1 is 2 kg, the mass of the box 2 is 4 kg, acceleration of gravity is 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup>, the magnitude of the force F is 40 Newton. The coefficient of the kinetic friction between the box 1 with the floor is 0.2 and the coefficient of the kinetic friction between the box 2 and floor is 0.3. Find (a) The magnitude and direction of the box’s acceleration<\/a> (b) Magnitude of the force exerted by the box 1 on the box 2 (F<\/span><\/span>12<\/span><\/sub>) and the magnitude of the force exerted by the box 2 on the box 1 (F<\/span><\/span>21<\/span><\/sub>).<\/span><\/span><\/p>\n

\"Motion<\/p>\n

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

\"Motion<\/p>\n

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

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

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

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

The force F = 40 Newton,<\/span><\/span><\/p>\n

Coefficient of the kinetic friction<\/a> between the box 1 with floor (<\/span><\/span>\u03bc<\/span><\/span>k1<\/span><\/sub>) <\/span><\/span>= 0.2<\/span><\/span><\/p>\n

Coefficient of the kinetic friction between the box 2 with floor (<\/span><\/span>\u03bc<\/span><\/span>k2<\/span><\/sub>) <\/span><\/span>= 0.3<\/span><\/span><\/p>\n

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

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

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

The normal force exerted on the box 2 (N<\/span><\/span>2<\/span><\/sub>) = w<\/span><\/span>2<\/span><\/sub> = 40 Newton<\/span><\/span><\/p>\n

The force of the kinetic friction exerted on the box 1 (f<\/span><\/span>k1<\/span><\/sub>) = (<\/span><\/span>\u03bc<\/span><\/span>k1<\/span><\/sub>)(N<\/span><\/span>1<\/span><\/sub>) = (0.2)(20) = 4 Newton<\/span><\/span><\/p>\n

The force of the kinetic friction exerted on the box 2 (f<\/span><\/span>k2<\/span><\/sub>) = (<\/span><\/span>\u03bc<\/span><\/span>k1<\/span><\/sub>)(N<\/span><\/span>2<\/span><\/sub>) = (0.3)(40) = 12 Newton<\/span><\/span><\/p>\n

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

(a) Magnitude and direction of the box’s acceleration<\/span><\/span><\/p>\n

\u03a3F = m a<\/span><\/span><\/p>\n

F – <\/span><\/span>f<\/span><\/span>k1<\/span><\/sub> – f<\/span><\/span>k2 <\/span><\/sub>= (m<\/span><\/span>1<\/span><\/sub> + m<\/span><\/span>2<\/span><\/sub>) a<\/span><\/span><\/p>\n

40 \u2013 4 \u2013 12 = (2 + 4) a<\/span><\/span><\/p>\n

24 = 6 a<\/span><\/span><\/p>\n

a = 24 \/ 6<\/span><\/span><\/p>\n

a = 4 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Direction of the acceleration = direction of the net force = rightward.<\/span><\/span><\/p>\n

(b) Magnitude of the force exerted by the box 1 on the box 2 (F<\/span><\/span>12<\/span><\/sub>) and the magnitude of the force exerted by the box 2 on the box 1 (F<\/span><\/span>21<\/span><\/sub>).<\/span><\/span><\/p>\n

Calculate the magnitude of F<\/span><\/span>12<\/span><\/sub> :<\/span><\/span><\/p>\n

\u03a3F = m a<\/span><\/span><\/p>\n

F<\/span><\/span>12<\/span><\/sub> – f<\/span><\/span>k2 <\/span><\/sub>= (m<\/span><\/span>2<\/span><\/span><\/sub>) a<\/span><\/span><\/p>\n

F<\/span><\/span>12<\/span><\/sub> – 12 = (4)(4)<\/span><\/span><\/p>\n

F<\/span><\/span>12<\/span><\/sub> – 12 = 16<\/span><\/span><\/p>\n

F<\/span><\/span>12<\/span><\/sub> = 16 + 12<\/span><\/span><\/p>\n

F<\/span><\/span>12<\/span><\/sub> = 28 Newton<\/span><\/span><\/p>\n

F<\/span><\/span>12<\/span><\/sub> and F<\/span><\/span>21 <\/span><\/sub>are action and reaction forces that act on the different objects. F<\/span><\/span>12<\/span><\/sub> and F<\/span><\/span>21<\/span><\/sub> has the same magnitude and opposite direction. <\/span><\/span><\/p>\n

F<\/span><\/span>12<\/span><\/sub> = 28 Newton = F<\/span><\/span>21<\/span><\/sub> = 28 Newton.<\/span><\/span><\/p>\n

<\/p>\n

2. Mass of the box 1 is 2 kg, mass of the box 2 is 4 kg, acceleration of gravity is 10 m\/s<\/span><\/span>2<\/span><\/span><\/sup>, the force F is 40 N. The coefficient of the kinetic friction between the box 1 with the floor is 0.2 and the coefficient of the kinetic friction between box 2 and floor is 0.3. Determine (a) Magnitude and direction of the acceleration (b) The tension in the cord connecting the boxes. Ignore cord’s mass.<\/span><\/span><\/p>\n

\"Motion<\/p>\n

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

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

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

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

The force F = 40 Newton,<\/span><\/span><\/p>\n

Coefficient of the kinetic friction between the box 1 with floor is 0.2 (<\/span><\/span>\u03bc<\/span><\/span>k1<\/span><\/sub>) <\/span><\/span>= 0.2<\/span><\/span><\/p>\n

Coefficient of the kinetic friction between the box 2 with floor is 0.2 (<\/span><\/span>\u03bc<\/span><\/span>k2<\/span><\/sub>) <\/span><\/span>= 0.3<\/span><\/span><\/p>\n

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

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

The normal force exerted on the box 1 (N<\/span><\/span>1<\/span><\/sub>) = w<\/span><\/span>1<\/span><\/sub> = 20 Newton<\/span><\/span><\/p>\n

The normal force exerted on the box 2 (N<\/span><\/span>2<\/span><\/sub>) = w<\/span><\/span>2<\/span><\/sub> = 40 Newton<\/span><\/span><\/p>\n

The force of the kinetic friction exerted on the box 1 (f<\/span><\/span>k1<\/span><\/sub>) = (<\/span><\/span>\u03bc<\/span><\/span>k1<\/span><\/sub>)(N<\/span><\/span>1<\/span><\/sub>) = (0.2)(20) = 4 Newton<\/span><\/span><\/p>\n

The force of the kinetic friction exerted on the box 2 (f<\/span><\/span>k2<\/span><\/sub>) = (<\/span><\/span>\u03bc<\/span><\/span>k1<\/span><\/sub>)(N<\/span><\/span>2<\/span><\/sub>) = (0.3)(40) = 12 Newton<\/span><\/span><\/p>\n

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

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

\u03a3F = m a<\/span><\/span><\/p>\n

F – <\/span><\/span>f<\/span><\/span>k1<\/span><\/sub> – f<\/span><\/span>k2 <\/span><\/sub>= (m<\/span><\/span>1<\/span><\/span><\/sub> + m<\/span><\/span>2<\/span><\/span><\/sub>) a<\/span><\/span><\/p>\n

40 \u2013 4 \u2013 12 = (2 + 4) a<\/span><\/span><\/p>\n

24 = 6 a<\/span><\/span><\/p>\n

a = 24 \/ 6<\/span><\/span><\/p>\n

a = 4 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Magnitude of the acceleration is 4 m\/s<\/span><\/span>2<\/span><\/span><\/sup>, direction of the acceleration = direction of the net force = rightward.<\/span><\/span><\/p>\n

(b) Tension in the cord<\/span><\/span><\/p>\n

Forces acts on the box 1 in the horizontal direction are the tension 1 (T<\/span><\/span>1<\/span><\/sub>) rightward and force of the kinetic friction 1 (f<\/span><\/span>k1<\/span><\/sub>) leftward. Apply Newton’s second law : <\/span><\/span><\/p>\n

\u03a3F = m a<\/span><\/span><\/p>\n

T<\/span><\/span>1<\/span><\/sub> – f<\/span><\/span>k1 <\/span><\/sub>= m<\/span><\/span>1<\/span><\/sub> a<\/span><\/span><\/p>\n

T<\/span><\/span>1<\/span><\/sub> – 4<\/span><\/span> = (2)(4)<\/span><\/span><\/p>\n

T<\/span><\/span>1<\/span><\/sub> – 4<\/span><\/span> = 8<\/span><\/span><\/p>\n

T<\/span><\/span>1 <\/span><\/sub>= 8 + 4 = 12 Newton<\/span><\/span><\/p>\n

The forces acts on the box 2 in the horizontal direction are the tension 2 (T<\/span><\/span>2<\/span><\/sub>) leftward and force of the kinetic friction 2 (f<\/span><\/span>k2<\/span><\/sub>) rightward. Apply Newton’s second law<\/a> : <\/span><\/span><\/p>\n

\u03a3F = m a<\/span><\/span><\/p>\n

F – T<\/span><\/span>2<\/span><\/sub> – f<\/span><\/span>k2 <\/span><\/sub>= m<\/span><\/span>2<\/span><\/sub> a<\/span><\/span><\/p>\n

40 \u2013 T<\/span><\/span>2<\/span><\/sub> \u2013 12 = (4)(4)<\/span><\/span><\/p>\n

28 – T<\/span><\/span>2<\/span><\/sub> = 16<\/span><\/span><\/p>\n

T<\/span><\/span>2 <\/span><\/sub>= 28 \u2013 16 = 12 Newton<\/span><\/span><\/p>\n

The tension in the cord connecting the boxes = T<\/span><\/span>1 <\/span><\/sub>= T<\/span><\/span>2<\/span><\/sub> = T = 12 Newton.<\/span><\/span><\/p>\n

<\/p>\n

[wpdm_package id=’493′]<\/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. Motion of two bodies with the same acceleration on rough horizontal surface with friction force<\/a><\/li>\n
  7. Motion on inclined plane without friction force<\/a><\/li>\n
  8. Motion on rough inclined plane with friction force<\/a><\/li>\n
  9. Motion in an elevator<\/a><\/li>\n
  10. 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. Mass of the box 1 is 2 kg, the mass of the box 2 is 4 kg, acceleration of gravity is 10 m\/s2, the magnitude of the force F is 40 Newton. The coefficient of the kinetic friction between the box 1 with the floor is 0.2 and the coefficient of the kinetic friction … 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":"Motion of two bodies with the same accelerations on the rough horizontal surface with the friction force - 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-195","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\/195","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=195"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/195\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=195"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=195"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=195"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}