{"id":1259,"date":"2018-02-28T15:34:57","date_gmt":"2018-02-28T07:34:57","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1259"},"modified":"2018-02-28T15:34:57","modified_gmt":"2018-02-28T07:34:57","slug":"the-magnitude-of-net-torque-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/the-magnitude-of-net-torque-problems-and-solutions.htm","title":{"rendered":"The magnitude of net torque \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. A force P is applied to one end of a beam with a length of 2 m. What is the magnitude of the torque<\/a>? The axis of rotation at point A. <\/span><\/span><\/p>\n

\"TheKnown :<\/u><\/span><\/span><\/p>\n

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

Length of AB (r<\/span><\/span>AB<\/span><\/sub>) = 2 m<\/span><\/span><\/p>\n

Force F is perpendicular to the beam. <\/i><\/span><\/span><\/p>\n

The lever arm (l<\/span><\/span>) <\/span><\/span>= r<\/span><\/span>AB <\/span><\/sub>sin 90<\/span><\/span>o <\/span><\/span><\/sup>= (2 m)(1) = 2 m <\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> The torque about the axis of rotation<\/span><\/span><\/p>\n

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

The torque :<\/span><\/span><\/p>\n

\u03c4<\/span><\/span> = F l = (10 N)(2 m) = 20 N m<\/span><\/span><\/p>\n

The plus sign because the beam <\/i><\/span><\/span>rotates counterclockwise rotation.<\/i><\/span><\/span><\/span> <\/i><\/p>\n\n

2. The length of a beam <\/span><\/span>AB is 2 m and the magnitude of force F is 10 N. What is the magnitude of the torque? The axis of rotation at point A. <\/span><\/span><\/p>\n

\"TheKnown :<\/u><\/span><\/span><\/p>\n

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

Length of AB (r<\/span><\/span>AB<\/span><\/sub>) = 2 m<\/span><\/span><\/p>\n

The lever arm (l<\/span><\/span>) <\/span><\/span>= r<\/span><\/span>AB <\/span><\/sub>sin 60<\/span><\/span>o <\/span><\/span><\/sup>= (2 m)(0.5<\/span><\/span>\u221a3<\/span><\/span>) = <\/span><\/span>\u221a3 <\/span><\/span>m<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> The torque about the axis of rotation<\/span><\/span><\/p>\n

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

The torque :<\/span><\/span><\/p>\n

\u03c4<\/span><\/span> = F l = (10 N)(<\/span><\/span>\u221a3 <\/span><\/span>m) = 10<\/span><\/span>\u221a3 <\/span><\/span>N m<\/span><\/span><\/p>\n

The plus sign because the force F causes the beam <\/i><\/span><\/span>rotates counterclockwise rotation.<\/i><\/span><\/span><\/span> <\/i><\/p>\n

<\/p>\n

3. The length of a beam is 2 m. The magnitude of F<\/span><\/span>1<\/span><\/span><\/sub> is 10 N and the magnitude of F<\/span><\/span>2<\/span><\/span><\/sub> is 15 N. Determine the net torque about the center of the beam. <\/span><\/span><\/p>\n

The axis of rotation at the center of the beam.<\/i><\/span><\/span><\/p>\n

\"TheKnown :<\/u><\/span><\/span><\/p>\n

Force 1 (F<\/span><\/span>1<\/span><\/sub>) = 10 N<\/span><\/span><\/p>\n

The distance<\/a> between F<\/span><\/span>1<\/span><\/span><\/sub> and the center of beam (r<\/span><\/span>1<\/span><\/span><\/sub>) = 1 m <\/span><\/span><\/p>\n

The lever arm 1 (l<\/span><\/span>1<\/span><\/span><\/sub>) <\/span><\/span>= r<\/span><\/span>1 <\/span><\/span><\/sub>sin 90<\/span><\/span>o <\/span><\/span><\/sup>= (1 m)(1) = 1 m<\/span><\/span><\/p>\n

Force F<\/i><\/span><\/span>1<\/i><\/span><\/sub> is perpendicular to the beam. <\/i><\/span><\/span><\/p>\n

Force 2 (F<\/span><\/span>2<\/span><\/sub>) = 15 N<\/span><\/span><\/p>\n

The distance between F<\/span><\/span>2<\/span><\/span><\/sub> and the center of the beam (r<\/span><\/span>2<\/span><\/sub>) = 1 m<\/span><\/span><\/p>\n

Force F<\/i><\/span><\/span>2<\/i><\/span><\/sub> is perpendicular to the beam. <\/i><\/span><\/span><\/p>\n

The lever arm 2 (l<\/span><\/span>2<\/span><\/sub>) <\/span><\/span>= r<\/span><\/span>1 <\/span><\/span><\/sub>sin 90<\/span><\/span>o <\/span><\/span><\/sup>= (1 m)(1) = 1 m<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> T<\/span><\/span>he net torque about the axis of rotation<\/span><\/span><\/p>\n

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

The torque 1 : <\/span><\/span><\/p>\n

\u03c4<\/span><\/span>1<\/span><\/span><\/sub> = F<\/span><\/span>1<\/span><\/span><\/sub> l<\/span><\/span>1<\/span><\/span><\/sub> = (10 N)(<\/span><\/span>1 <\/span><\/span>m) = 10<\/span><\/span> N m<\/span><\/span><\/p>\n

The plus sign because the force of F<\/i><\/span><\/span>1 <\/i><\/span><\/span><\/sub>causes the beam <\/i><\/span><\/span>rotates counterclockwise rotation.<\/i><\/span><\/span><\/span> <\/i><\/p>\n

The torque 2 : <\/span><\/span><\/p>\n

\u03c4<\/span><\/span>2<\/span><\/span><\/sub> = F<\/span><\/span>2<\/span><\/sub> l<\/span><\/span>2 <\/span><\/sub>= (15 N)(<\/span><\/span>1 <\/span><\/span>m) = -15 N <\/span><\/span><\/p>\n

The minus sign because the force F<\/i><\/span><\/span>2<\/i><\/span><\/span><\/sub> causes the beam to <\/i><\/span><\/span>rotates clockwise.<\/i><\/span><\/span><\/span> <\/i><\/p>\n

The net torque :<\/span><\/span><\/p>\n

\u03a3<\/span><\/span>\u03c4 = \u03c4<\/span><\/span>1 <\/span><\/span><\/sub>\u2013 \u03c4<\/span><\/span>2 <\/span><\/span><\/sub>= 10 \u2013 15 = – 5 N m <\/span><\/span><\/p>\n

The minus sign because the beam to <\/i><\/span><\/span>rotates clockwise.<\/i><\/span><\/span><\/span> <\/i><\/p>\n

<\/p>\n

4. The length beam AB is <\/span><\/span>2 m, The magnitude of F<\/span><\/span>1<\/span><\/span><\/sub> is 10 N and the magnitude of F<\/span><\/span>2 <\/span><\/span><\/sub>is <\/span><\/span>10 N. Determine the net torque about the center of the beam. <\/span><\/span><\/p>\n

\"TheThe axis of rotation at the center of the beam.<\/i><\/span><\/span><\/p>\n

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

Force 1 (F<\/span><\/span>1<\/span><\/span><\/sub>) = 10 Nn<\/span><\/span><\/p>\n

The distance between F<\/span><\/span>1<\/span><\/span><\/sub> and the center of the beam (r<\/span><\/span>1<\/span><\/span><\/sub>) = 1 m<\/span><\/span><\/p>\n

The lever arm 1 (l<\/span><\/span>1<\/span><\/span><\/sub>) <\/span><\/span>= r<\/span><\/span>1 <\/span><\/span><\/sub>sin 60<\/span><\/span>o <\/span><\/span><\/sup>= (1 m)(0.5<\/span><\/span>\u221a3<\/span><\/span>) = 0.5<\/span><\/span>\u221a3 <\/span><\/span>m<\/span><\/span><\/p>\n

Force 2 (F<\/span><\/span>2<\/span><\/span><\/sub>) = 10 N<\/span><\/span><\/p>\n

The distance between F<\/span><\/span>2<\/span><\/span><\/sub> and the center of the beam (r<\/span><\/span>2<\/span><\/span><\/sub>) = 1 m<\/span><\/span><\/p>\n

Force F<\/i><\/span><\/span>2<\/i><\/span><\/span><\/sub> is perpendicular to the beam. <\/i><\/span><\/span><\/p>\n

The lever arm 2 (l<\/span><\/span>2<\/span><\/span><\/sub>) = r<\/span><\/span>2 <\/span><\/span><\/sub>sin 90<\/span><\/span>o <\/span><\/span><\/sup>= (1 m)(1) = 1 m<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> The net torque about the axis of rotation<\/span><\/span><\/p>\n

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

The torque 1 :<\/span><\/span><\/p>\n

\u03c4<\/span><\/span>1<\/span><\/span><\/sub> = F<\/span><\/span>1<\/span><\/span><\/sub> l<\/span><\/span>1<\/span><\/span><\/sub> = (10 N)(<\/span><\/span>0.5<\/span><\/span>\u221a3 <\/span><\/span>m) = <\/span><\/span>5<\/span><\/span>\u221a3 = 8.7 N.m<\/span><\/span><\/p>\n

The plus sign because the force F<\/i><\/span><\/span>1<\/i><\/span><\/span><\/sub> causes the beam to <\/i><\/span><\/span>rotates clockwise.<\/i><\/span><\/span><\/span> <\/i><\/p>\n

The torque 2 :<\/span><\/span><\/p>\n

\u03c4<\/span><\/span>2<\/span><\/span><\/sub> = F<\/span><\/span>2<\/span><\/span><\/sub> l<\/span><\/span>2 <\/span><\/span><\/sub>= (10 N)(<\/span><\/span>1 <\/span><\/span>m) = -10 N m<\/span><\/span><\/p>\n

The minus sign because the force F<\/i><\/span><\/span>2<\/i><\/span><\/span><\/sub> causes the beam to <\/i><\/span><\/span>rotates clockwise.<\/i><\/span><\/span><\/span><\/p>\n

The net torque :<\/span><\/span><\/p>\n

\u03a3<\/span><\/span>\u03c4 = \u03c4<\/span><\/span>1 <\/span><\/span><\/sub>\u2013 \u03c4<\/span><\/span>2 <\/span><\/span><\/sub>= 8.7 \u2013 10 = – 1.3 N m<\/span><\/span><\/p>\n

The minus sign because the net force causes the beam to <\/i><\/span><\/span>rotates clockwise.<\/i><\/span><\/span><\/span> <\/i><\/p>\n

<\/p>\n

5. The length of a beam is 10 m, the magnitude of F<\/span><\/span>1 <\/span><\/span><\/sub>is 10 N, the magnitude of F<\/span><\/span>2<\/span><\/span><\/sub> is 10 N and the magnitude of F<\/span><\/span>3<\/span><\/span><\/sub> is 15 N. The distance between point A and point C is 7.5 m. The Force F<\/span><\/span>2<\/span><\/span><\/sub> located at the center of the beam. Determine the net torque about the point C located at 2.5 m from the point B. <\/span><\/span><\/p>\n

The axis of rotation located at point C<\/i><\/span><\/span><\/p>\n

\"TheKnown :<\/u><\/span><\/span><\/p>\n

Force 1 (F<\/span><\/span>1<\/span><\/sub>) = 10 N<\/span><\/span><\/p>\n

The distance between F<\/span><\/span>1<\/span><\/span><\/sub> and point C (r<\/span><\/span>1<\/span><\/span><\/sub>) = <\/span><\/span>2.5 <\/span><\/span>m<\/span><\/span><\/p>\n

Force F<\/i><\/span><\/span>1<\/i><\/span><\/sub> is perpendicular to the beam. <\/i><\/span><\/span><\/p>\n

The lever arm 1 (l<\/span><\/span>1<\/span><\/sub>) <\/span><\/span>= r<\/span><\/span>1 <\/span><\/sub>sin 90<\/span><\/span>o <\/span><\/span><\/sup>= (<\/span><\/span>2.5 <\/span><\/span>m)(<\/span><\/span>1<\/span><\/span>) = <\/span><\/span>2.5<\/span><\/span> m<\/span><\/span><\/p>\n

Force 2 (F<\/span><\/span>2<\/span><\/sub>) = 10 N<\/span><\/span><\/p>\n

The distance between F<\/span><\/span>2<\/span><\/sub> and point C (r<\/span><\/span>2<\/span><\/sub>) = <\/span><\/span>2.5 <\/span><\/span>m<\/span><\/span><\/p>\n

The force F<\/i><\/span><\/span>2<\/i><\/span><\/sub> is perpendicular to the beam. <\/i><\/span><\/span><\/p>\n

The lever arm 2 (l<\/span><\/span>2<\/span><\/sub>) = r<\/span><\/span>2 <\/span><\/sub>sin 90<\/span><\/span>o <\/span><\/span><\/sup>= (<\/span><\/span>2.5 <\/span><\/span>m)(1) = <\/span><\/span>2.5 <\/span><\/span>m<\/span><\/span><\/p>\n

Force 3 (F<\/span><\/span>3<\/span><\/sub>) = 15 N<\/span><\/span><\/p>\n

The distance between F<\/span><\/span>3<\/span><\/span><\/sub> and point <\/span><\/span>C (r<\/span><\/span>3<\/span><\/sub>) = 7.5 m<\/span><\/span><\/p>\n

The force F<\/i><\/span><\/span>3<\/i><\/span><\/sub> is perpendicular to the beam. <\/i><\/span><\/span><\/p>\n

The lever arm 3 (l<\/span><\/span>3<\/span><\/sub>) = r<\/span><\/span>3<\/span><\/sub> sin 90<\/span><\/span>o <\/span><\/span><\/sup>= (7.5 m)(1) = 7.5 m<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> The net torque about the axis of rotation<\/span><\/span><\/p>\n

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

The torque 1 :<\/span><\/span><\/p>\n

\u03c4<\/span><\/span>1<\/span><\/sub> = F<\/span><\/span>1<\/span><\/sub> l<\/span><\/span>1<\/span><\/sub> = (10 N)(<\/span><\/span>2.5 <\/span><\/span>m) = <\/span><\/span>25 <\/span><\/span>N m<\/span><\/span><\/p>\n

The plus sign because the force F<\/i><\/span><\/span>1<\/i><\/span><\/span><\/sub> causes the beam to <\/i><\/span><\/span>rotates clockwise.<\/i><\/span><\/span><\/span> <\/i><\/p>\n

The torque 2 :<\/span><\/span><\/p>\n

\u03c4<\/span><\/span>2<\/span><\/sub> = F<\/span><\/span>2<\/span><\/sub> l<\/span><\/span>2 <\/span><\/sub>= (10 N)(<\/span><\/span>2.5<\/span><\/span> m) = <\/span><\/span>25 N m<\/span><\/span><\/p>\n

The plus sign because the force F<\/i><\/span><\/span>2<\/i><\/span><\/sub> causes the beam to <\/i><\/span><\/span>rotates clockwise.<\/i><\/span><\/span><\/span> <\/i><\/p>\n

The torque 3 :<\/span><\/span><\/p>\n

\u03c4<\/span><\/span>3<\/span><\/sub> = F<\/span><\/span>3<\/span><\/sub> l<\/span><\/span>3<\/span><\/sub> = (15 N)(<\/span><\/span>7.5 <\/span><\/span>m) = -112..5<\/span><\/span> N m<\/span><\/span><\/p>\n

The minus sign because the force F<\/i><\/span><\/span>3<\/i><\/span><\/span><\/sub> causes the beam to <\/i><\/span><\/span>rotates clockwise.<\/i><\/span><\/span><\/span> <\/i><\/p>\n

The net torque :<\/span><\/span><\/p>\n

\u03a3<\/span><\/span>\u03c4 = \u03c4<\/span><\/span>1 <\/span><\/sub>+ \u03c4<\/span><\/span>2 <\/span><\/sub>– \u03c4<\/span><\/span>3 <\/span><\/sub>= <\/span><\/span>25 <\/span><\/span>+ <\/span><\/span>25 \u2013 112.5 <\/span><\/span>= – 62.5 N m<\/span><\/span><\/p>\n

The minus sign because the net force causes the beam to <\/i><\/span><\/span>rotates clockwise.<\/i><\/span><\/span><\/span> <\/i><\/p>\n

<\/p>\n

6. The length of a beam is 10 m, the magnitude of F<\/span><\/span>1<\/span><\/sub> is 10 N, the magnitude of F<\/span><\/span>2<\/span><\/sub> is 10 N and the magnitude of F<\/span><\/span>3<\/span><\/sub> is 10 N. Determine the net torque about point A, located 5 m from the poi\"Thent of application of force F<\/span><\/span>1<\/span><\/span><\/sub>. <\/span><\/span><\/p>\n

The axis of rotation at point A.<\/i><\/span><\/span><\/p>\n

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

Force 1 (F<\/span><\/span>1<\/span><\/span><\/sub>) = 10 N<\/span><\/span><\/p>\n

The distance between F<\/span><\/span>1<\/span><\/sub> and point A (r<\/span><\/span>1<\/span><\/sub>) = 5 m <\/span><\/span><\/p>\n

The lever arm 1 (l<\/span><\/span>1<\/span><\/sub>) <\/span><\/span>= r<\/span><\/span>1 <\/span><\/span><\/sub>sin 60<\/span><\/span>o <\/span><\/span><\/sup>= (5 m)(0.5<\/span><\/span>\u221a3<\/span><\/span>) = 2.5<\/span><\/span>\u221a3 <\/span><\/span>m <\/span><\/span><\/p>\n

Force<\/a> 2 (F<\/span><\/span>2<\/span><\/sub>) = 10 N<\/span><\/span><\/p>\n

The distance between F<\/span><\/span>2<\/span><\/sub> and point A (r<\/span><\/span>2<\/span><\/sub>) = 0 <\/span><\/span><\/p>\n

The force F<\/i><\/span><\/span>2<\/i><\/span><\/span><\/sub> is perpendicular to the beam.<\/i><\/span><\/span><\/p>\n

The lever arm 2 (l<\/span><\/span>2<\/span><\/sub>) = r<\/span><\/span>2 <\/span><\/span><\/sub>sin 90<\/span><\/span>o <\/span><\/span><\/sup>= (0)(1) = 0 <\/span><\/span><\/p>\n

The force 3 (F<\/span><\/span>3<\/span><\/sub>) = 10 N<\/span><\/span><\/p>\n

The distance between F<\/span><\/span>3<\/span><\/sub> and point A (r<\/span><\/span>3<\/span><\/sub>) = 10 m <\/span><\/span><\/p>\n

The lever arm 3 (l<\/span><\/span>3<\/span><\/sub>) = r<\/span><\/span>3 <\/span><\/sub>sin 30<\/span><\/span>o <\/span><\/span><\/sup>= (10 m)(0.5) = 5 m <\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> the net torque about the axis of rotation<\/span><\/span><\/p>\n

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

The torque 1 :<\/span><\/span><\/p>\n

\u03c4<\/span><\/span>1<\/span><\/sub> = F<\/span><\/span>1<\/span><\/sub> l<\/span><\/span>1<\/span><\/sub> = (10 N)(<\/span><\/span>2.5<\/span><\/span>\u221a3 m<\/span><\/span>) = <\/span><\/span>25\u221a3 = 43.3 <\/span><\/span>N m<\/span><\/span><\/p>\n

The plus sign because the force F<\/i><\/span><\/span>1<\/i><\/span><\/sub> causes the beam to <\/i><\/span><\/span>rotates clockwise.<\/i><\/span><\/span><\/span> <\/i><\/p>\n

The torque 2 :<\/span><\/span><\/p>\n

\u03c4<\/span><\/span>2<\/span><\/sub> = F<\/span><\/span>2<\/span><\/sub> l<\/span><\/span>2 <\/span><\/sub>= (10 N)(0) = 0<\/span><\/span><\/p>\n

The torque 3 :<\/span><\/span><\/p>\n

\u03c4<\/span><\/span>3<\/span><\/sub> = F<\/span><\/span>3<\/span><\/sub> l<\/span><\/span>3<\/span><\/sub> = (10 N)(<\/span><\/span>5 <\/span><\/span>m) = -50<\/span><\/span> N m<\/span><\/span><\/p>\n

The minus sign because the force F<\/i><\/span><\/span>3<\/i><\/span><\/span><\/sub> causes the beam to <\/i><\/span><\/span>rotates clockwise.<\/i><\/span><\/span><\/span> <\/i><\/p>\n

The net torque :<\/span><\/span><\/p>\n

\u03a3<\/span><\/span>\u03c4 = \u03c4<\/span><\/span>1 <\/span><\/sub>+ \u03c4<\/span><\/span>2 <\/span><\/sub>– \u03c4<\/span><\/span>3 <\/span><\/sub>= <\/span><\/span>43.3 <\/span><\/span>+ 0<\/span><\/span> \u2013 50 <\/span><\/span>= – 6.7 N m<\/span><\/span><\/p>\n

The minus sign because the net force causes the beam to <\/span><\/span>rotates clockwise.<\/span><\/span><\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

1. A force P is applied to one end of a beam with a length of 2 m. What is the magnitude of the torque? The axis of rotation at point A. Known : Force (F) = 10 N Length of AB (rAB) = 2 m Force F is perpendicular to the beam. The lever … 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":"The magnitude of net torque \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-1259","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\/1259","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=1259"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1259\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1259"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1259"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1259"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}