{"id":8130,"date":"2023-04-30T12:32:21","date_gmt":"2023-04-30T12:32:21","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=8130"},"modified":"2024-05-25T06:52:24","modified_gmt":"2024-05-25T06:52:24","slug":"rope-tension-equation","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/rope-tension-equation.htm","title":{"rendered":"Rope tension equation","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

3 Questions about Rope tension equation<\/span><\/p>\n

1. The picture below shows three blocks, namely A, B and C which are located on a smooth horizontal plane. If mass A = 1 kg, mass B = 2 kg and mass C = 2 kg and F = 10 N, then determine the ratio of the tension in the rope between A and B to the tension in the rope between B and C.<\/span><\/span><\/p>\n

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

The mass of A (mA<\/sub>) = 1 kg<\/span><\/p>\n

Mass B (mB<\/sub>) = 2 kg<\/span><\/p>\n

The mass of C (mC<\/sub>) = 2 kg<\/span><\/p>\n

Tensile force (F) = 10 N<\/span><\/p>\n

Wanted:<\/u> TAB<\/sub> : TBC<\/sub><\/span><\/p>\n

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

Calculate the acceleration of the system using Newton’s Second Law formula:<\/span><\/p>\n

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

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

10 = (1 + 2 + 2) a<\/span><\/p>\n

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

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

a = 2 m\/s2<\/sup><\/span><\/p>\n

Use the rope tension formula to calculate TAB<\/sub><\/span><\/p>\n

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

TAB <\/sub>= mA<\/sub> a = 1 (2) = 2 Newton<\/span><\/p>\n

Use the rope tension formula to calculate TBC<\/sub><\/span><\/p>\n

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

TBC<\/sub> = (mA<\/sub> + mB<\/sub>) a = (1 + 2) (2) = (3)(2) = 6 Newton<\/span><\/p>\n

2. Object A with a mass of 6 kg and object B with a mass of 3 kg are connected by a rope as shown. If the coefficient of friction is 0.3 and g = 10 m\/s2<\/sup>, determine the acceleration of the object and the tension in the ropes of each block.<\/span><\/p>\n

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

The mass of object A (mA<\/sub>) = 6 kg<\/span><\/p>\n

The mass of object B (mB<\/sub>) = 3 kg<\/span><\/p>\n

Coefficient of friction of block A (\u00b5k<\/sub>) = 0.3<\/span><\/p>\n

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

The weight of block A (wA<\/sub>) = mA<\/sub> g = (6)(10) = 60 N<\/span><\/p>\n

Normal force on block A (NA<\/sub>) = wA<\/sub> = 60 N<\/span><\/p>\n

The weight of block B (wB<\/sub>) = mB<\/sub> g = (3)(10) = 30 N<\/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

Calculate the kinetic frictional force i.e. the frictional force when block A moves:<\/span><\/p>\n

Fk<\/sub> = \u00b5k<\/sub> NA <\/sub>= (0,3)(60) = 18 Newton<\/span><\/p>\n

Calculate the acceleration of the system (a):<\/b><\/span><\/p>\n

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

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

30 \u2013 18 = (6 + 3) a<\/span><\/p>\n

12 = 9 a<\/span><\/p>\n

a = 12 \/ 9 = 1,3 m\/s2<\/sup><\/span><\/p>\n

Calculate the tension in the string on block A (TA<\/sub>):<\/b><\/span><\/p>\n

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

TA<\/sub> – Fk<\/sub> = (mA<\/sub>) a<\/span><\/p>\n

TA<\/sub> \u2013 18 = (6)(1,3)<\/span><\/p>\n

TA<\/sub> \u2013 18 = 7,8<\/span><\/p>\n

TA<\/sub> = 7,8 + 18 = 25,8 Newton <\/span><\/p>\n

Calculate the tension in the rope on beam B (TB<\/sub>):<\/b><\/span><\/p>\n

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

wB<\/sub> \u2013 TB <\/sub>= mB<\/sub> (a)<\/span><\/p>\n

30 – TB <\/sub>= 3 (1,3)<\/span><\/p>\n

30 – TB <\/sub>= 3,9<\/span><\/p>\n

TB <\/sub>= 30 \u2013 3,9<\/span><\/p>\n

TB <\/sub>= 26,1 Newton<\/span><\/p>\n

3. Two objects A and B with masses of 5 kg and 3 kg are connected by a frictionless pulley. The force P is applied to the pulley in an upward direction. If both blocks are initially at rest on the floor, what is the acceleration of block A, if the magnitude of P is 60 N?<\/span><\/p>\n

Determine also the tension in the rope on blocks A and B.<\/span><\/p>\n

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

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

The mass of A (mA<\/sub>) = 5 kg<\/span><\/p>\n

The weight of block A (wA<\/sub>) = mA<\/sub> g = (5)(10) = 50 Newton<\/span><\/p>\n

Mass B (mB<\/sub>) = 3 kg<\/span><\/p>\n

The weight of block B (wB<\/sub>) = mB<\/sub> g = (3)(10) = 30 Newton<\/span><\/p>\n

Force P = 60 N<\/span><\/p>\n

Wanted:<\/u> Acceleration of the system of beams A and B (a) and the tension in the rope on beam A (TA<\/sub>) and beam B (TB<\/sub>)<\/span><\/p>\n

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

Calculate the acceleration of the system using Newton’s Second Law formula.<\/b><\/span><\/p>\n

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

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

50 \u2013 30 = (5 + 3) a<\/span><\/p>\n

20 = 8 a<\/span><\/p>\n

a = 20 \/ 8<\/span><\/p>\n

a = 2,5 m\/s2<\/sup><\/span><\/p>\n

Use the tension force formula to calculate the tension in the rope<\/b><\/span><\/p>\n

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

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

wA<\/sub> \u2013 TA<\/sub> = mA<\/sub> a<\/span><\/p>\n

50 \u2013 TA <\/sub>= 5 (2,5)<\/span><\/p>\n

50 \u2013 TA <\/sub>= 12,5<\/span><\/p>\n

TA<\/sub> = 50 \u2013 12,5 = 37,5 Newton<\/span><\/p>\n

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

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

TB<\/sub> \u2013 wB<\/sub> = mB<\/sub> a<\/span><\/p>\n

TB<\/sub> \u2013 30 = 3 (2,5)<\/span><\/p>\n

TB<\/sub> \u2013 30 = 7,5<\/span><\/p>\n

TB<\/sub> = 7,5 + 30 = 37,5 Newton<\/span><\/p>\n

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

3 Questions about Rope tension equation 1. The picture below shows three blocks, namely A, B and C which are located on a smooth horizontal plane. If mass A = 1 kg, mass B = 2 kg and mass C = 2 kg and F = 10 N, then determine the ratio of the tension … 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":"Rope tension 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-8130","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\/8130","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=8130"}],"version-history":[{"count":7,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/8130\/revisions"}],"predecessor-version":[{"id":9851,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/8130\/revisions\/9851"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=8130"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=8130"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=8130"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}