{"id":154,"date":"2018-01-17T11:09:28","date_gmt":"2018-01-17T03:09:28","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=154"},"modified":"2018-01-17T11:09:28","modified_gmt":"2018-01-17T03:09:28","slug":"normal-force-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/normal-force-problems-and-solutions.htm","title":{"rendered":"Normal force \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Solved problems in Newton’s laws of motion – Normal force\u00a0<\/a><\/p>\n

1. An object resting on a table, shown in the figure below. Mass of the object is 1 kg. Acceleration of gravity<\/a> is 9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>. Determine the normal force exerted on the object by the table.<\/span><\/span><\/p>\n

\"Normal-force-\u2013-problems-and-solutions-1-1\"<\/p>\n

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

Mass (m) = 1 kg<\/span><\/span><\/p>\n

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

Weight (w) = m g = (1 kg)(9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>) = 9.8 kg m\/s<\/span><\/span>2<\/span><\/span><\/sup> = 9.8 Newton<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> normal force (N)<\/span><\/span><\/p>\n

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

\"Normal<\/p>\n

The object is at rest on the table, so the net force on the object is zero (Newton’s first or second law). The weight of the object acts vertically downward, toward the center of the Earth. There must be another force on the object to balance the gravitational force<\/a>. Object resting on the table, so that the table exerts this upward force. The force exerted by the table is often called a normal force (N). Normal means perpendicular. <\/span><\/span><\/p>\n

Choose the upward direction as the positive y-direction. The net force on the object is :<\/span><\/span><\/p>\n

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

N \u2013 w = 0<\/span><\/span><\/p>\n

N = w<\/span><\/span><\/p>\n

N = m g<\/span><\/span><\/p>\n

N = 9.8 Newton<\/span><\/span><\/p>\n

The normal force on the object, exerted by the table is 9.8 N upward.<\/span><\/span><\/p>\n

<\/p>\n

2. Two objects resting on a table. Mass<\/a> of object 1 (m<\/span><\/span>1<\/span><\/span><\/sub>) = 1 kg, mass of object 2 (m<\/span><\/span>2<\/span><\/span><\/sub>) = 2 kg, acceleration due to gravity (g) =9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>. Determine the magnitude and direction of the normal force exerted by m<\/span><\/span>2<\/span><\/span><\/sub> on the m<\/span><\/span>1<\/span><\/span><\/sub> and the normal force exerted by the table on the m<\/span><\/span>2<\/span><\/span><\/sub>. <\/span><\/span><\/p>\n

\"Normal<\/p>\n

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

\"Normal<\/p>\n

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

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

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

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

Weight<\/a> of object 1 (w<\/span><\/span>1<\/span><\/sub>) = m<\/span><\/span>1<\/span><\/sub> g = (1)(9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>) = 9.8 kg m\/s<\/span><\/span>2<\/span><\/span><\/sup> = 9.8 Newton<\/span><\/span><\/p>\n

Weight of object 2 (w<\/span><\/span>2<\/span><\/sub>) = m<\/span><\/span>2<\/span><\/sub> g = (2)(9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>) = 19.6 kg m\/s<\/span><\/span>2<\/span><\/span><\/sup> = 19.6 Newton<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> N<\/span><\/span>1<\/span><\/sub> and N<\/span><\/span>2<\/span><\/sub><\/p>\n

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

(a) Normal force exerted by m<\/span><\/span>2<\/span><\/sub> to the m<\/span><\/span>1<\/span><\/sub> (N<\/span><\/span>1<\/span><\/sub>)<\/span><\/span><\/p>\n

N<\/span><\/span>1<\/span><\/sub> = w<\/span><\/span>1<\/span><\/sub> = 9.8 Newton<\/span><\/span><\/p>\n

Direction of N<\/span><\/span>1<\/span><\/sub> is upward.<\/span><\/span><\/p>\n

(b) Normal force exerted by the table on the m<\/span><\/span>2<\/span><\/sub> (N<\/span><\/span>2<\/span><\/sub>)<\/span><\/span><\/p>\n

N<\/span><\/span>2<\/span><\/sub> = w<\/span><\/span>1<\/span><\/sub> + w<\/span><\/span>2<\/span><\/sub> = 9.8 Newton + 19.6 Newton = 29.4 Newton<\/span><\/span><\/p>\n

Direction of N<\/span><\/span>2<\/span><\/sub> is upward.<\/span><\/span><\/p>\n

<\/p>\n

3. An object resting on the table. Mass of the object is 2 kg, acceleration due to gravity is 9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>. Magnitude of the force F is 10 Newton. Find the magnitude and direction of the normal force exerted by the table on the object. <\/span><\/span><\/p>\n

\"Normal<\/p>\n

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

\"Normal<\/p>\n

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

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

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

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

Force F (F) = 10 Newton<\/span><\/span><\/p>\n

Wanted<\/u><\/span><\/span> : magnitude and direction of the normal force (N)<\/span><\/span><\/p>\n

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

direction of the normal force is upward.<\/span><\/span><\/p>\n

Magnitude of the normal force :<\/span><\/span><\/p>\n

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

N \u2013 F – w = 0<\/span><\/span><\/p>\n

N = F + w<\/span><\/span><\/p>\n

N = 10 Newton + 20 Newton <\/span><\/span><\/p>\n

N = 30 Newton<\/span><\/span><\/p>\n

<\/p>\n

4. An object resting on a table. Object’s mass is 1 kg, acceleration due to gravity is 9,8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>, force F<\/span><\/span>1<\/span><\/sub> is 10 N and force F<\/span><\/span>2<\/span><\/sub> is 20 N. Determine magnitude and direction of the normal force exerted by the table on the object. g = 9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

\"Normal<\/p>\n

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

\"Normal<\/p>\n

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

Mass (m) = 1 kg<\/span><\/span><\/p>\n

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

Weight (w) = m g = (1 kg)(9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>) = 9.8 kg m\/s<\/span><\/span>2<\/span><\/span><\/sup> = 9.8 Newton<\/span><\/span><\/p>\n

F<\/span><\/span>1<\/span><\/sub> = 10 Newton<\/span><\/span><\/p>\n

F<\/span><\/span>2<\/span><\/sub> = 20 Newton<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> magnitude and direction of the normal force (N)<\/span><\/span><\/p>\n

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

Direction of the normal force is upward.<\/span><\/span><\/p>\n

Magnitude of the normal force :<\/span><\/span><\/p>\n

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

N \u2013 F<\/span><\/span>2<\/span><\/sub> \u2013 w + F<\/span><\/span>1<\/span><\/sub> = 0<\/span><\/span><\/p>\n

N = F<\/span><\/span>2<\/span><\/sub> + w – F<\/span><\/span>1<\/span><\/sub><\/p>\n

N = 20 Newton + 9.8 Newton – 10 Newton <\/span><\/span><\/p>\n

N = 19.8 Newton<\/span><\/span><\/p>\n

<\/p>\n

5. Object’s mass (m) = 2 kg, acceleration of gravity (g) = 9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>, angle = 30<\/span><\/span>o<\/span><\/span><\/sup>. Find magnitude and direction of the normal force exerted on the object.<\/span><\/span><\/p>\n

\"Normal<\/p>\n

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

\"Normal<\/p>\n

w is weight, w<\/span><\/span>x<\/span><\/sub> is horizontal component of the weight, w<\/span><\/span>y<\/span><\/sub> is a vertical component of the weight, N is the normal force.<\/span><\/span><\/p>\n

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

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

acceleration of gravity (g) = 9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

weight (w) = m g = (2 kg)(9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>) = 19.6 kg m\/s<\/span><\/span>2<\/span><\/span><\/sup> = 19.6 Newton<\/span><\/span><\/p>\n

w<\/span><\/span>x<\/span><\/sub> = w sin 60<\/span><\/span>o<\/span><\/span><\/sup> = (19.6 N)(0.5)<\/span><\/span>\u221a<\/span><\/span>3<\/span><\/span>= 9.8<\/span><\/span>\u221a<\/span><\/span>3 <\/span><\/span>Newton<\/span><\/span><\/p>\n

w<\/span><\/span>y<\/span><\/sub> = w cos 60 = (19.6 N)(0.5) = 9.8 <\/span><\/span>Newton<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> normal force <\/span><\/span>(N)<\/span><\/span><\/p>\n

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

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

N \u2013 w<\/span><\/span>y<\/span><\/sub> = 0<\/span><\/span><\/p>\n

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

N = 9.8 Newton<\/span><\/span><\/p>\n

<\/p>\n

[wpdm_package id=’467′]<\/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":"

    Solved problems in Newton’s laws of motion – Normal force\u00a0 1. An object resting on a table, shown in the figure below. Mass of the object is 1 kg. Acceleration of gravity is 9.8 m\/s2. Determine the normal force exerted on the object by the table. Known : Mass (m) = 1 kg Acceleration of … 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":"Normal force \u2013 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-154","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\/154","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=154"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/154\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=154"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=154"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=154"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}