{"id":3795,"date":"2023-10-05T19:01:18","date_gmt":"2023-10-05T19:01:18","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=3795"},"modified":"2023-10-05T19:03:00","modified_gmt":"2023-10-05T19:03:00","slug":"week-7-ss3-first-term-further-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-7-ss3-first-term-further-mathematics-notes\/","title":{"rendered":"Week 7 – SS3 First Term Further Mathematics Notes"},"content":{"rendered":"

WEEK SEVEN
\n<\/strong>STATICS \u2013 CONTINUATION<\/strong>
\n\t\tDefinition of Moment of a Force
\nPrinciples of Moments
\nApplication of the Principle in Solving Problems<\/p>\n

\u00a0Definition
\n\t\t\t\t<\/strong>Moments of a Force<\/strong>: The moment of a force about a reference point is defined as the product of the force and the force arm. Moment of a force is a vector quantity and its units is Nm.<\/p>\n

\u00a0The direction of movement or sense of movement about the point is duly considered.
\nThe object can move in clockwise moment and ant-clockwise moments about the given point.<\/p>\n

\u00a0Suppose we have an object acted upon by two forces, F1 and F2<\/sub> in the opposite direction, then
\n\"\"\/\"\"\/\"\"\/\"\"\/\"\"\/\"\"\/\"\"\/\"\"\/M1<\/sub> = F1<\/sub> x di M2<\/sub>= F2<\/sub> x d2<\/sub>
\n\t\tWhere M1<\/sub> = Magnitude of F1<\/sub> M2<\/sub> = Magnitude of F2<\/sub>
\n\t\t d1 = Force arm of d1<\/sub> d2<\/sub> = Force arm of d2<\/sub><\/p>\n

\u00a0PRINCIPLE OF MOMENTS
\n1.\u00a0\u00a0\u00a0\u00a0When a system of coplanar forces is in equilibrium, then the sum of the clockwise moment is equal to the sum of the anti-clockwise moments about the same point in the plane.
\n2.\u00a0\u00a0\u00a0\u00a0When the system is not in equilibrium, then the resultant of two coplanar forces F1<\/sub> and F2<\/sub> denote by R is represented by the relationship below:
\n M1<\/sub> + M2<\/sub> + MR
\n<\/sub>
\n\u00a0Where M1<\/sub> = Moment of F1 M2<\/sub> = Moments of F2<\/sub> MR<\/sub> = Moments of T.<\/p>\n

\u00a0Centre of Gravity<\/strong>: The centre of gravity of a uniform plank or rod is the midpoint of the plant or rod\"\"\/\"\"\/\"\"\/\"\"\/\"\"\/<\/p>\n

\u00a0
\n\u00a0
\n\u00a0Examples
\n<\/strong>1. A uniform rod PQ is 15m long and has mass 20kg. The rod rests on two supports at P and Q. An object of mass 5kg is suspended at a point R on the rod 5m from the end P. Calculate the reaction of the supports P and Q (take g = 10ms-2<\/sup>)
\nSolution\"\"\/\"\"\/\"\"\/\"\"\/\"\"\/\"\"\/
\n\t\t\u00a0\u00a0\u00a0\u00a0Kpkq
\n\"\"\/
\n\t\t\"\"\/\u00a0\u00a0\u00a0\u00a07.5m\u00a0\u00a0\u00a0\u00a0
\n\"\"\/\u00a0\u00a0\u00a0\u00a0 5m
\n\"\"\/\"\"\/ P Q
\n\u00a0\u00a0\u00a0\u00a0
\n\u00a0\u00a0\u00a0\u00a0\u00a0
\n\u00a0 5kg 20kg
\nLet the reaction at P be Kp and at Q beKq<\/sub>
\n\t\tNB: The weight of the rod acts downward through the midpoint of the rod.<\/p>\n

\u00a0Moments about the point P, MR<\/sub> = M1<\/sub> + M2<\/sub>
\n\t\tBut M1<\/sub> = F1<\/sub> x d1<\/sub> and F = Mg.
\nKq x 15 = (5 x 10 ) x 5 + ( 20 x 10 ) x 7.5
\n15kq = 50 x 5 + 200 x 7.5
\nKq<\/sub> = 250 + 1500
\n 15.
\nKq = 116. 7N.
\nMoment about the point Q,
\nKp x 15 = ( 5 x 10 ) x 10 + (20 x 10 ) x 7.5.
\n15Kp = 500 + 1500
\nKp = 2000
\n 15.
\nKp = 133.3N<\/p>\n

    \n
  1. A uniform beam AB of length 6m and mass 20kg rests on support P and Q placed 1m from each end of the beam. Masses of 10kg and 8kg are placed at A and B respectively. Calculate the reactions at P and Q (g = 9.8ms-2<\/sup>)\n<\/li>\n<\/ol>\n

    \u00a0Solution
    \n<\/strong>\"\"\/\"\"\/ 8kg
    \n10kg
    \n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
    \n\"\"\/\"\"\/\"\"\/
    \n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
    \n 20kg <\/p>\n

    \u00a0Let reaction at point P be Rp and at point Q be Rq.
    \n:. Moment about the point Q
    \nRp x 4 = ( 10 x 9.8) x 5 + (20 x 9.8 ) x 2 \u2013 ( 8 x 9.8 x 1 )
    \n4Rp = 490 + 392 – 78.4
    \nRp = 803.6
    \n 4. Rp = 200.9N<\/p>\n

    \u00a0Moment about the point P:
    \nRq x 4 = ( 8 x 9.8 ) x 5 + (20 x 9.8 ) x 2 – ( 10 x 9.8) x 1
    \n4Rq = 392 + 392 \u2013 98
    \nRq = 689
    \n 4
    \nRq = 171.5N<\/p>\n

    \u00a0EVALUATION
    \n\t\t\t\t<\/em><\/strong>A uniform rod PQ, is 20 m long and weighs 80N, has weights 20 N and 50N suspended at P and Q respectively. Find the distance from P where the rod must be supported so that it will rest horizontally. <\/p>\n

    \u00a0\"\"\/GENERAL EVALUATION
    \n<\/strong>\"\"\/\"\"\/\"\"\/A uniform rod PQ of length 10 m and mass 2kg rest on two supports at x and y. If PX = 2m and QY = 1m, find the reaction of X. (Take g = 10ms-2<\/sup>)
    \nWEEKEND ASSIGNMENT
    \n<\/strong>A uniform beam PQ of length 100cm and of weight 35N lies on a support 40cm from the end P. Weights 54N and W are attached to the ends P and Q respectively to keep the beam in equilibrium. Find the value of W, to the nearest whole number.<\/p>\n

    \u00a0READING ASIGNMENT<\/strong>:
    \nRead Rotational Equilibrium and Principle of Moments. Page 178-185 of Further Mathematics Project III.<\/p>\n","protected":false},"excerpt":{"rendered":"

    WEEK SEVEN STATICS \u2013 CONTINUATION Definition of Moment of a Force Principles of Moments Application…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,294],"tags":[],"class_list":["post-3795","post","type-post","status-publish","format-standard","hentry","category-posts","category-first-term-ss3-further-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3795","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=3795"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3795\/revisions"}],"predecessor-version":[{"id":3796,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3795\/revisions\/3796"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=3795"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=3795"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=3795"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}