{"id":206,"date":"2018-01-19T08:10:34","date_gmt":"2018-01-19T00:10:34","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=206"},"modified":"2018-01-19T08:10:34","modified_gmt":"2018-01-19T00:10:34","slug":"motion-on-rough-inclined-plane-with-friction-force-application-of-newtons-law-of-motion-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/motion-on-rough-inclined-plane-with-friction-force-application-of-newtons-law-of-motion-problems-and-solutions.htm","title":{"rendered":"Motion on the rough inclined plane with the friction force – application of Newton’s law of motion problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. Object’s mass<\/a> = 2 kg, acceleration due to gravity<\/a> = 9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>, coefficient of the static friction<\/a> = 0.2, coefficient of the kinetic friction = 0.1. Is the object at rest or accelerating? If the object is accelerated, find (a) the net force (b) magnitude and direction of the box’s acceleration<\/a>!<\/span><\/span><\/p>\n

\"Motion<\/p>\n

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

\"Motion<\/p>\n

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

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

Acceleration due to gravity (g) = 9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Coefficient of the static friction (<\/span><\/span>\u03bc<\/span><\/span>s<\/span><\/sub>) <\/span><\/span>= 0.2<\/span><\/span><\/p>\n

Coefficient of the kinetic friction (<\/span><\/span>\u03bc<\/span><\/span>k<\/span><\/sub>)<\/span><\/span> = 0.1<\/span><\/span><\/p>\n

Weight (w) = m g = (2)(9.8) = <\/span><\/span>19.6 Newton<\/span><\/span><\/p>\n

The horizontal component of the weight<\/a> (w<\/span><\/span>x<\/span><\/sub>)<\/span><\/span> = w sin 30<\/span><\/span>o<\/span><\/span><\/sup> = (19.6)(0.5) = 9.8 Newton <\/span><\/span><\/p>\n

The vertical component of th weight (w<\/span><\/span>y<\/span><\/sub>)<\/span><\/span> = w cos 30<\/span><\/span>o<\/span><\/span><\/sup> = (19.6)(0.5\u221a3) = 9.8\u221a3 Newton<\/span><\/span><\/p>\n

The normal force<\/a> (N) = w<\/span><\/span>y<\/span><\/sub> = 9.8\u221a3 Newton<\/span><\/span><\/p>\n

Force of the static friction (f<\/span><\/span>s<\/span><\/sub>) = (0.2)(9.8\u221a3) = 1.96\u221a3 Newton = 3.39 Newton <\/span><\/span><\/p>\n

Force of the kinetic friction (f<\/span><\/span>k<\/span><\/sub>) = (0.1)(9.8\u221a3) = 0.98\u221a3 Newton = 1.69 Newton <\/span><\/span><\/p>\n

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

Object is at rest if w<\/span><\/span>x<\/span><\/sub> < f<\/span><\/span>s<\/span><\/sub>, object is moving down if w<\/span><\/span>x<\/span><\/sub> > f<\/span><\/span>s<\/span><\/sub>. <\/span><\/span><\/p>\n

wx<\/span><\/sub> = 9.8 Newton and f<\/span><\/span>s <\/span><\/sub>= 3.39 Newton.<\/span><\/span><\/p>\n

(a) the net force<\/u><\/span><\/span><\/p>\n

\u2211F = w<\/span><\/span>x <\/span><\/sub>\u2013 f<\/span><\/span>k<\/span><\/sub> = 9.8 \u2013 1.69 = 8.11 Newton<\/span><\/span><\/p>\n

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

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

8.11 = (2) a<\/span><\/span><\/p>\n

a = 4.05 <\/span><\/span><\/p>\n

Magnitude of the acceleration = 4.05 m\/s<\/span><\/span>2 <\/span><\/span><\/sup>and direction of the acceleration = downward.<\/span><\/span><\/p>\n

<\/p>\n

2. Object’s mass = 4 kg, acceleration due to gravity = 9,8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>. Coefficient of the kinetic friction = 0.2 and coefficient of the static friction = 0.4. Magnitude of the force F = 40 Newton. The object is at rest or slides down ? If the object slides down, find (a) the net force (b) magnitude and direction of the acceleration! <\/span><\/span><\/p>\n

\"Motion<\/p>\n

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

\"Motion<\/p>\n

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

Mass (m) = 4 kg<\/span><\/span><\/p>\n

Acceleration due to gravity (g) = 9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

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

The coefficient of the kinetic friction <\/span><\/span>(<\/span><\/span>\u03bc<\/span><\/span>k<\/span><\/sub>)<\/span><\/span> = 0.2<\/span><\/span><\/p>\n

Weight (w) = m g = (4)(9.8) = 39.2 Newton<\/span><\/span><\/p>\n

The horizontal component of the weight (w<\/span><\/span>x<\/span><\/sub>) <\/span><\/span>= w sin 30<\/span><\/span>o<\/span><\/span><\/sup> = (39.2)(0.5) = 19.6 Newton <\/span><\/span><\/p>\n

The vertical component of the weight (w<\/span><\/span>y<\/span><\/sub>)<\/span><\/span> = w cos 30<\/span><\/span>o<\/span><\/span><\/sup> = (392)(0..5\u221a3) = 19.6\u221a3 Newton<\/span><\/span><\/p>\n

The normal force (N) = w<\/span><\/span>y<\/span><\/sub> = 19.6\u221a3 Newton = 33.95 Newton<\/span><\/span><\/p>\n

the static friction force (f<\/span><\/span>s<\/span><\/sub>) = <\/span><\/span>\u03bc<\/span><\/span>s <\/span><\/sub>N = <\/span><\/span>(0,4)(33.95) = 13.58 Newton <\/span><\/span><\/p>\n

The kinetic friction force (f<\/span><\/span>k<\/span><\/sub>) = <\/span><\/span>\u03bc<\/span><\/span>k <\/span><\/sub>N = <\/span><\/span>(0.2)(33.95) = 6.79 Newton <\/span><\/span><\/p>\n

F = 40 Newton<\/span><\/span><\/p>\n

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

The object slides down if F < w<\/span><\/span>x<\/span><\/sub> + f<\/span><\/span>s<\/span><\/sub>. The object slides up if F > w<\/span><\/span>x<\/span><\/sub> + f<\/span><\/span>s<\/span><\/sub>. <\/span><\/span><\/p>\n

F = 40 Newton, w<\/span><\/span>x<\/span><\/sub> = 19.6 Newton and f<\/span><\/span>s <\/span><\/sub>= 13.58 Newton. <\/span><\/span><\/p>\n

F is greater than w<\/span><\/span>x<\/span><\/sub> + f<\/span><\/span>s <\/span><\/sub>so the object slides up.<\/span><\/span><\/p>\n

(a) The net force<\/u><\/span><\/span><\/p>\n

\u2211F = F – w<\/span><\/span>x<\/span><\/sub> – <\/span><\/span>f<\/span><\/span>k<\/span><\/sub> = 40 \u2013 19.6 – 6.79 = 13.61 Newton<\/span><\/span><\/p>\n

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

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

6.4 = (4) a<\/span><\/span><\/p>\n

a = 1.6 <\/span><\/span><\/p>\n

The magnitude of the acceleration is 1.6 m\/s<\/span><\/span>2 <\/span><\/span><\/sup>and direction of the acceleration is upward<\/span><\/span>.<\/span><\/span><\/p>\n

<\/p>\n

[wpdm_package id=’481′]<\/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 the 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. Object’s mass = 2 kg, acceleration due to gravity = 9.8 m\/s2, coefficient of the static friction = 0.2, coefficient of the kinetic friction = 0.1. Is the object at rest or accelerating? If the object is accelerated, find (a) the net force (b) magnitude and direction of the box’s acceleration! Solution Known : … 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 on the rough inclined plane with the friction force - 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-206","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\/206","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=206"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/206\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=206"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=206"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=206"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}