{"id":2156,"date":"2018-04-25T16:32:17","date_gmt":"2018-04-25T08:32:17","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=2156"},"modified":"2023-08-09T01:31:50","modified_gmt":"2023-08-09T01:31:50","slug":"moment-of-force-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/moment-of-force-problems-and-solutions.htm","title":{"rendered":"Moment of force \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Moment of force \u2013 problems and solutions<\/p>\n

1. If FR<\/sub> is the net force of F1<\/sub>, F2<\/sub>, and F3<\/sub>, what is the magnitude of force F2<\/sub> and x?<\/span><\/span><\/p>\n

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

Net force (FR<\/sub>) = 40 N\"Moment<\/span><\/span><\/p>\n

Force 1 (F1<\/sub>) = 10 N<\/span><\/span><\/p>\n

Force (F3<\/sub>) = 20 N<\/span><\/span><\/p>\n

Wanted:<\/u> The magnitude of force F2<\/sub> and distance of x<\/span><\/span><\/p>\n

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

Find the magnitude of force F2 <\/sub>:<\/span><\/span><\/p>\n

Force points to upward, signed negative and force points to downward, signed negative.<\/i><\/span><\/span><\/p>\n

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

– FR<\/sub> + F1<\/sub> + F2<\/sub> \u2013 F3<\/sub> = 0<\/span><\/span><\/p>\n

– 40 + 10 + F2 <\/sub>\u2013 20 = 0<\/span><\/span><\/p>\n

– 30 + F2 <\/sub>\u2013 20 = 0<\/span><\/span><\/p>\n

– 50 + F2 <\/sub>= 0<\/span><\/span><\/p>\n

F2 <\/sub>= 50 Newton.<\/span><\/span><\/p>\n

Plus sign indicates that the direction of the force is upward.<\/span><\/span><\/p>\n

Find x.<\/span><\/span><\/p>\n

Choose A as the axis of rotation.<\/span><\/span><\/p>\n

\u03c41 <\/sub>= F1<\/sub> l1<\/sub> = (10 N)(1 m) = 10 Nm <\/span><\/span><\/p>\n

The torque 1 rotates beam counterclockwise so we assign positive sign to the torque 3.<\/i><\/span><\/span><\/p>\n

\u03c42 <\/sub>= F2<\/sub> x = (50)(x) = 50x Nm<\/span><\/span><\/p>\n

The torque 1 rotates beam counterclockwise so we assign positive sign to the torque 3.<\/i><\/span><\/span><\/p>\n

\u03c43 <\/sub>= F3<\/sub> x = (20 N)(1.75 m) = -35 Nm<\/span><\/span><\/p>\n

The torque 2 rotates beam clockwise so we assign negative sign to the torque 2.<\/i><\/span><\/span><\/p>\n

The net of moment of force<\/a> :<\/span><\/span><\/p>\n

\u03a3\u03c4 = 0 <\/span><\/span><\/p>\n

10 + 50x \u2013 35 = 0<\/span><\/span><\/p>\n

50x – 25 = 0<\/span><\/span><\/p>\n

50x = 25<\/span><\/span><\/p>\n

x = 25\/50<\/span><\/span><\/p>\n

x = 0.5 m<\/span><\/span><\/p>\n

2. Forces of F1<\/sub>, F2<\/sub>, F3<\/sub>, and F4<\/sub> acts on the rod of ABCD as shown in figure. If rod’s mass ignored, what is the magnitude of the moment of force, about point A.<\/span><\/span><\/p>\n

The axis of rotation = points A.<\/span><\/span><\/p>\n

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

Force F1 <\/sub>= 10 N, the lever arm l1<\/sub> = 0 \"Moment<\/span><\/span><\/p>\n

Force F2 <\/sub>= 4 N, the lever arm l2<\/sub> = 2 meters<\/span><\/span><\/p>\n

Force F3 <\/sub>= 5 N, the lever arm l3<\/sub> = 3 meters<\/span><\/span><\/p>\n

Force F4 <\/sub>= 10 N, the lever arm l4<\/sub> = 6 meters<\/span><\/span><\/p>\n

Wanted :<\/u> the moment of force about point A<\/span><\/span><\/p>\n

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

Moment of force 1 (\u03c41<\/sub>) = F1<\/sub> l1<\/sub> = (10)(0) = 0<\/span><\/span><\/p>\n

Moment of force 2 (\u03c42<\/sub>) = F2<\/sub> l2<\/sub> = (4)(2) = -8 Nm<\/span><\/span><\/p>\n

Moment of force 3 (\u03c43<\/sub>) = F3<\/sub> l3<\/sub> = (5)(3) = 15 Nm<\/span><\/span><\/p>\n

Moment of force 4 (\u03c44<\/sub>) = F4<\/sub> l4<\/sub> = (10)(6) = -60 Nm<\/span><\/span><\/p>\n

If torque rotates rod counterclockwise then we assign positive sign. <\/i><\/span><\/span><\/p>\n

If torque rotates rod clockwise then we assign negative sign.<\/i><\/span><\/span><\/p>\n

The resultant of the moment of force :<\/span><\/span><\/p>\n

\u03c4 = 0 – 8 Nm + 15 Nm – 60 Nm<\/span><\/span><\/p>\n

\u03c4 = -68 Nm + 15 Nm <\/span><\/span><\/p>\n

\u03c4 = -53 Nm <\/span><\/span><\/p>\n

Minus sign indicates that the moment of force rotates rod clockwise.<\/span><\/span><\/p>\n

3. Three forces act on a rod, FA<\/sub> = FC<\/sub> = 10 N and FB<\/sub> = 20 N, as shown in figure below. If distance of AB = BC = 20 cm, what is the moment of force about point C.<\/span><\/span><\/p>\n

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

The axis rotation at point C.\"Moment<\/span><\/span><\/p>\n

Distance between FA<\/sub> and the axis of rotation (rAC<\/sub>) = 40 cm = 0,4 meters<\/span><\/span><\/p>\n

Distance between FB<\/sub> and the axis of rotation (rBC<\/sub>) = 20 cm = 0.2 meters<\/span><\/span><\/p>\n

Distance between FC<\/sub> and the axis of rotation (rCC<\/sub>) = 0 cm<\/span><\/span><\/p>\n

FA<\/sub> = 10 Newton<\/span><\/span><\/p>\n

FB<\/sub> = 20 Newton<\/span><\/span><\/p>\n

FC<\/sub> = 10 Newton<\/span><\/span><\/p>\n

Wanted :<\/u> The resultant of the moment of force about point C.<\/span><\/span><\/p>\n

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

Moment of force A :<\/u><\/span><\/span><\/p>\n

\u03a3\u03c4A <\/sub>= (FA<\/sub>)(rAC<\/sub> sin 90o<\/sup>) = (10 N)(0,4 m)(1) = -4 N.m<\/span><\/span><\/p>\n

Minus sign indicates that the moment of force rotates rod clockwise.<\/i><\/span><\/span><\/p>\n

Moment of force B :<\/u><\/span><\/span><\/p>\n

\u03a3\u03c4B <\/sub> = (FB<\/sub>)(rBC<\/sub> sin 90o<\/sup>) = (20 N)(0,2 m)(1) = 4 N.m<\/span><\/span><\/p>\n

Plus sign indicates that the moment of force rotates rod counterclockwise.<\/i><\/span><\/span><\/p>\n

Moment of force C :<\/u><\/span><\/span><\/p>\n

\u03a3\u03c4C <\/sub>= (FC<\/sub>)(rCC<\/sub> sin 90o<\/sup>) = (10 N)(0)(1) = 0<\/span><\/span><\/p>\n

The resultant of the moment of force :<\/u><\/span><\/span><\/p>\n

\u03a3\u03c4 = \u03a3\u03c41<\/sub> + \u03a3\u03c42<\/sub> + \u03a3\u03c43<\/sub> <\/span><\/span><\/p>\n

\u03a3\u03c4 = -4 + 4 + 0<\/span><\/span><\/p>\n

\u03a3\u03c4 = 0 N.m<\/span><\/span><\/p>\n

4. Length of a rod is 50 cm. Three forces act on the rod, as shown in figure below. If the axis of rotation is point C, what is the net of the moment of force.<\/span><\/span><\/p>\n

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

The axis rotation at point C.\"Moment<\/span><\/span><\/p>\n

Distance between F1<\/sub> and the axis of rotation is (r1<\/sub>) = 30 cm = 0,3 meters<\/span><\/span><\/p>\n

Distance between F2<\/sub> and the axis of rotation (r2<\/sub>) = 10 cm = 0,1 meters<\/span><\/span><\/p>\n

Distance between F3<\/sub> and the axis of rotation (r3<\/sub>) = 20 cm = 0,2 meters<\/span><\/span><\/p>\n

F1<\/sub> = 10 Newton<\/span><\/span><\/p>\n

F2<\/sub> = 10 Newton<\/span><\/span><\/p>\n

F3<\/sub> = 10 Newton<\/span><\/span><\/p>\n

Wanted :<\/u> Resultant of moment of force about point C.<\/span><\/span><\/p>\n

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

Moment of force 1 :<\/u><\/span><\/span><\/p>\n

\u03a3\u03c41<\/sub> = (F1<\/sub>)(r1<\/sub> sin 90o<\/sup>) = (10 N)(0,3 m)(1) = -3 N.m<\/span><\/span><\/p>\n

Minus sign indicates that the moment of force rotates rod clockwise.<\/i><\/span><\/span><\/p>\n

Moment of force 2 :<\/u><\/span><\/span><\/p>\n

\u03a3\u03c42 <\/sub>= (F2<\/sub>)(r2<\/sub> sin 90o<\/sup>) = (10 N)(0,1 m)(1) = 1 N.m<\/span><\/span><\/p>\n

Plus sign indicates that the moment of force rotates rod counterclockwise.<\/i><\/span><\/span><\/p>\n

Moment of force 3 :<\/u><\/span><\/span><\/p>\n

\u03a3\u03c43 <\/sub>= (F3<\/sub>)(r3<\/sub> sin 30o<\/sup>) = (10 N)(0,2 m)(0,5) = -1 N.m<\/span><\/span><\/p>\n

Minus sign indicates that the moment of force rotates rod clockwise.<\/i><\/span><\/span><\/p>\n

The resultant of the moment of force :<\/span><\/span><\/p>\n

\u03a3\u03c4 = \u03a3\u03c41 <\/sub> + \u03a3\u03c42<\/sub> + \u03a3\u03c43<\/sub> <\/span><\/span><\/p>\n

\u03a3\u03c4 = -3 + 1 – 1<\/span><\/span><\/p>\n

\u03a3\u03c4 = -3 N.m<\/span><\/span><\/p>\n

Minus sign indicates that the resultant of the moment of force rotates rod clockwise.<\/i><\/span><\/span><\/p>\n

5. Three forces F1<\/sub>, F2<\/sub>, and F3<\/sub> act on a rod as shown in figure below. Length of rod is 4 meters. What is the moment of force about point C.<\/span><\/span><\/p>\n

(sin 53o<\/sup> = 0.8, cos 53o<\/sup> = 0.6, AB = BC = CD = DE = 1 meter)<\/span><\/span><\/p>\n

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

The axis of rotation at point C. \"Moment<\/i><\/span><\/span><\/p>\n

Force 1 (F1<\/sub>) = 5 Newton<\/span><\/span><\/p>\n

The distance between the line of action of F1<\/sub> with the axis of rotation (r1<\/sub>) = 2 meters <\/i><\/span><\/span><\/p>\n

Force 2 (F2<\/sub>) = 0.4 Newton<\/span><\/span><\/p>\n

The distance between the line of action of F2<\/sub> with the axis of rotation (r2<\/sub>) = 1 meter <\/i><\/span><\/span><\/p>\n

Force 3 (F3<\/sub>) = 4.8 Newton<\/span><\/span><\/p>\n

The distance between the line of action of <\/i>F<\/i>3<\/i><\/sub> <\/i>with the axis of rotation <\/i>(r<\/i>3<\/i><\/sub>) = <\/i>2<\/i> meter <\/i><\/span><\/span><\/p>\n

Wanted:<\/u> The moment of force about point C.<\/span><\/span><\/p>\n

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

Moment of force 1 :<\/span><\/span><\/p>\n

\u03c41 <\/sub>= F1<\/sub> r sin 53o <\/sup>= (5 N)(2 m)(0,8) = (10)(0,8) N = 8 N <\/span><\/span><\/p>\n

Plus sign indicates that the moment of force rotates rod counterclockwise.<\/i><\/span><\/span><\/p>\n

Moment of force 2 :<\/span><\/span><\/p>\n

\u03c42 <\/sub>= F2<\/sub> r sin 90o <\/sup>= (0,4 N)(1 m)(1) = -0,4 N <\/span><\/span><\/p>\n

Minus sign indicates that the moment of force rotates rod clockwise.<\/i><\/span><\/span><\/p>\n

Moment of force 3 :<\/span><\/span><\/p>\n

\u03c43 <\/sub>= F3<\/sub> r sin 90o <\/sup>= (4,8 N)(2 m)(1) = -9,6 N <\/span><\/span><\/p>\n

Minus sign indicates that the moment of force rotates rod clockwise.<\/i><\/span><\/span><\/p>\n

The resultant of the moment of force :<\/span><\/span><\/p>\n

\u03a3\u03c4 = \u03c41 <\/sub> – \u03c42<\/sub> – \u03c43<\/sub> = 8 \u2013 0,4 \u2013 9,6 = 8 \u2013 10 = 2 N.m<\/span><\/span><\/p>\n

Plus sign indicates that the moment of force rotates rod counterclockwise.<\/i><\/span><\/span><\/p>\n

6. What is the resultant of the moment of force about the axis of rotation at point O by forces acts on the rod, as shown in the figure below?<\/span><\/span><\/p>\n

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

The axis of rotation at point O. \"Moment<\/i><\/span><\/span><\/p>\n

Force 1 (F1<\/sub>) = 6 Newton<\/span><\/span><\/p>\n

The distance between the line of action of F1<\/sub> with the axis of rotation (r1<\/sub>) = 1 meter <\/i><\/span><\/span><\/p>\n

Force 2 (F2<\/sub>) = 6 Newton<\/span><\/span><\/p>\n

The distance between the line of action of F2<\/sub> with the axis of rotation (r2<\/sub>) = 2 meters <\/i><\/span><\/span><\/p>\n

Force 3 (F3<\/sub>) = 4 Newton<\/span><\/span><\/p>\n

The distance between the line of action of F3<\/sub> with the axis of rotation (r3<\/sub>) = 2 meters <\/i><\/span><\/span><\/p>\n

Wanted:<\/u> The resultant of the moment of force about point C<\/span><\/span><\/p>\n

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

Moment of force 1 :<\/span><\/span><\/p>\n

\u03c41 <\/sub>= F1<\/sub> l1<\/sub> = (6 N)(1 m) = 6 Nm<\/span><\/span><\/p>\n

Plus sign indicates that the moment of force rotates rod counterclockwise.<\/i><\/span><\/span><\/p>\n

Moment of force 2 :<\/span><\/span><\/p>\n

\u03c42 <\/sub>= F2<\/sub> r2<\/sub> sin 30o <\/sup>= (6 N)(2 m)(0,5)= 6 Nm<\/span><\/span><\/p>\n

Plus sign indicates that the moment of force rotates rod counterclockwise.<\/i><\/span><\/span><\/p>\n

Moment of force 3 :<\/span><\/span><\/p>\n

\u03c43 <\/sub>= F3<\/sub> l3<\/sub> = (4 N)(2 m) = -8 Nm<\/span><\/span><\/p>\n

Minus sign indicates that the moment of force rotates rod clockwise.<\/i><\/span><\/span><\/p>\n

The resultant of the moment of force :<\/span><\/span><\/p>\n

\u03a3\u03c4 = \u03c41 <\/sub> + \u03c42<\/sub> – \u03c43<\/sub> = 6 + 6 – 8 = 4 N.m<\/span><\/span><\/p>\n

Plus sign indicates that the moment of force rotates rod counterclockwise.<\/i><\/span><\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

Moment of force \u2013 problems and solutions 1. If FR is the net force of F1, F2, and F3, what is the magnitude of force F2 and x? Known : Net force (FR) = 40 N Force 1 (F1) = 10 N Force (F3) = 20 N Wanted: The magnitude of force F2 and distance … 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":"","_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":"Moment of 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-2156","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\/2156","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=2156"}],"version-history":[{"count":2,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/2156\/revisions"}],"predecessor-version":[{"id":8636,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/2156\/revisions\/8636"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=2156"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=2156"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=2156"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}