{"id":8140,"date":"2023-04-30T21:45:09","date_gmt":"2023-04-30T21:45:09","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=8140"},"modified":"2024-05-25T06:51:39","modified_gmt":"2024-05-25T06:51:39","slug":"friction-force-equation","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/friction-force-equation.htm","title":{"rendered":"Friction force equation","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

3 questions about Friction force equation<\/span><\/p>\n

1. Block A 3 kg is placed on the table and then tied to a rope that is connected to stone B = 2 kg through a pulley as shown. The mass and friction of the pulleys are neglected. Acceleration due to gravity g = 10 m\/s2<\/sup>. Determine the acceleration of the system and the tension in the rope if:<\/span><\/p>\n

a) smooth table\"Friction<\/span><\/p>\n

b) rough table with a coefficient of kinetic friction of 0.4<\/span><\/p>\n

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

The mass of block A (mA<\/sub>) = 3 kg<\/span><\/p>\n

The mass of rock B (mB<\/sub>) = 2 kg<\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s2<\/sup><\/span><\/p>\n

Weight of block A (wA<\/sub>) = m g = (3)(10) = 30 Newton<\/span><\/p>\n

Weight of rock B (wB<\/sub>) = m g = (2)(10) = 20 Newton<\/span><\/p>\n

Wanted:<\/u> The acceleration of the system (a) and the tension in the rope (T)<\/span><\/p>\n

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

a) smooth table<\/i><\/span><\/p>\n

Calculate the acceleration of the system using the formula for Newton’s second law:<\/b><\/span><\/p>\n

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

wB<\/sub> = (mA<\/sub> + mB<\/sub>) a<\/span><\/p>\n

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

20 = 5 a<\/span><\/p>\n

a = 20 \/ 5 = 4 m\/s2<\/sup><\/span><\/p>\n

Calculate the tension in the rope using the formula for the tension in the rope:<\/b><\/span><\/p>\n

The tension in the rope on block A:<\/b><\/span><\/p>\n

\u03a3F = mA<\/sub> a<\/span><\/p>\n

T = mA<\/sub> a = (3)(4) = 12 Newton<\/span><\/p>\n

The tension in the rope on block B:<\/b><\/span><\/p>\n

\u03a3F = mB<\/sub> a<\/span><\/p>\n

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

20 \u2013 T = 8<\/span><\/p>\n

T = 20 \u2013 8 = 12 Newton <\/span><\/p>\n

b) rough table with a coefficient of kinetic friction of 0.4\"Friction<\/i><\/span><\/p>\n

The force of the kinetic Friction:<\/span><\/p>\n

Fk<\/sub> = \u00b5k<\/sub> N = (0,4)(30) = 12 Newton <\/span><\/p>\n

Calculate the acceleration of the system using the formula for Newton’s second law:<\/b><\/span><\/p>\n

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

wB<\/sub> \u2013 fk<\/sub> = (mA<\/sub> + mB<\/sub>) a<\/span><\/p>\n

20 \u2013 12 = (3 + 2) a<\/span><\/p>\n

8 = 5 a<\/span><\/p>\n

a = 8 \/ 5 = 1,6 m\/s2<\/sup><\/span><\/p>\n

Calculate the tension in the rope using the formula for the tension in the rope:<\/b><\/span><\/p>\n

The tension in the rope on block A:<\/span><\/p>\n

\u03a3F = mA<\/sub> a<\/span><\/p>\n

T \u2013 fk = mA<\/sub> a <\/span><\/p>\n

T \u2013 12 = (3)(1,6) <\/span><\/p>\n

T \u2013 12 = 4,8<\/span><\/p>\n

T = 4,8 + 12 = 16,8 Newton<\/span><\/p>\n

The tension in the rope on block B:<\/span><\/p>\n

\u03a3F = mB<\/sub> a<\/span><\/p>\n

wB<\/sub> \u2013 T = (2)(1,6)<\/span><\/p>\n

20 \u2013 T = 3,2<\/span><\/p>\n

T = 20 \u2013 3,2 = 16,8 Newton <\/span><\/p>\n

2. An object with a mass of 10 kg is in a horizontal plane. The coefficient of static friction is 0.4 and the coefficient of kinetic friction is 0.35. g = 10 m\/s2<\/sup>. If an object is given a constant horizontal force of 25 N, the magnitude of the frictional force acting on the object is…<\/span><\/p>\n

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

The mass of the object (m) = 10 kg<\/span><\/p>\n

The coefficient of Static friction (\u00b5s<\/sub>) = 0.4<\/span><\/p>\n

The coefficient of kinetic friction (\u00b5k) = 0.35<\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s2<\/sup><\/span><\/p>\n

Horizontal force (F) = 25 N<\/span><\/p>\n

The object’s gravity (w) = m g = (10)(10) = 100 Newton<\/span><\/p>\n

Normal force (N) = w = 100 Newton<\/span><\/p>\n

Wanted:<\/u> The amount of static friction (fs<\/sub>) and kinetic (fk<\/sub>)<\/span><\/p>\n

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

The force of the static Friction::<\/span><\/p>\n

fs<\/sub> = \u00b5s<\/sub> N = (0,4)(100) = 40 Newton <\/span><\/p>\n

The force of the Kinetic Friction:<\/span><\/p>\n

fk<\/sub> = \u00b5k<\/sub> N = (0,35)(100) = 35 Newton<\/span><\/p>\n

The horizontal force is only 25 Newton so it can’t move objects yet.<\/span><\/p>\n

3. The masses of blocks A and B in the figure are 10 kg and 5 kg respectively. The coefficient of friction between block A and the plane is 0.2. To prevent block A from moving, the minimum mass of block C required is…<\/span><\/p>\n

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

The mass of block A (mA<\/sub>) = 10 kg<\/span><\/p>\n

The mass of block B (mB<\/sub>) = 5 kg<\/span><\/p>\n

Coefficient of static friction of block A (\u00b5s<\/sub>) = 0,2<\/span><\/p>\n

Gravity acceleration (g) = 10 m\/s2<\/sup><\/span><\/p>\n

Block weight A (wA<\/sub>) = mA<\/sub> g = (10)(10) = 100 Newton <\/span><\/p>\n

Block weight B (wB<\/sub>) = mB<\/sub> g = (5)(10) = 50 Newton<\/span><\/p>\n

Static friction (fs<\/sub>) = \u00b5s<\/sub> N = (0,2)(wA<\/sub> + wC<\/sub>) = (0,2)(100 + wC<\/sub>) = 20 + 0,2 wC<\/sub> <\/span><\/p>\n

Ditanya:<\/u> The mass of block C to keep the system at rest<\/span><\/p>\n

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

The system is at rest so the formula for Newton’s first law is used:<\/span><\/p>\n

\u03a3F = 0<\/span><\/p>\n

wB<\/sub> \u2013 fs<\/sub> = 0<\/span><\/p>\n

50 \u2013 (20 + 0,2 wC<\/sub>) = 0 <\/span><\/p>\n

50 \u2013 20 – 0,2 wC<\/sub> = 0 <\/span><\/p>\n

30 – 0,2 wC<\/sub> = 0 <\/span><\/p>\n

30 = 0,2 wC<\/sub><\/span><\/p>\n

wC <\/sub>= 30 \/ 0,2 = 300 \/ 2 = 150 Newton<\/span><\/p>\n

The mass of block C = 150 \/ 10 = 15 Kg<\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

3 questions about Friction force equation 1. Block A 3 kg is placed on the table and then tied to a rope that is connected to stone B = 2 kg through a pulley as shown. The mass and friction of the pulleys are neglected. Acceleration due to gravity g = 10 m\/s2. Determine the … Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"closed","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":"3","_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":"Friction force equation","_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-8140","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\/8140","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=8140"}],"version-history":[{"count":7,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/8140\/revisions"}],"predecessor-version":[{"id":9849,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/8140\/revisions\/9849"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=8140"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=8140"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=8140"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}