{"id":3791,"date":"2023-10-05T18:59:07","date_gmt":"2023-10-05T18:59:07","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=3791"},"modified":"2023-10-05T19:03:00","modified_gmt":"2023-10-05T19:03:00","slug":"week-5-ss3-first-term-further-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-5-ss3-first-term-further-mathematics-notes\/","title":{"rendered":"Week 5 &#8211; SS3 First Term Further Mathematics  Notes"},"content":{"rendered":"<p>\u00a0<strong>WEEK FIVE<br \/>\n<\/strong><strong>STATICS \u2013 CONTINUATION<br \/>\n<\/strong>Definition of equilibrium<br \/>\nCondition of equilibrium of rigid body<br \/>\nApplication of the condition to solve problems<br \/>\nLami&#8217;s theorem and application<\/p>\n<p>\u00a0<strong>Definitions<br \/>\n<\/strong>Equilibrium is when a body remains at rest under the action of given forces.<\/p>\n<p>\u00a0<strong>Translational Equilibrium<\/strong>: The state of equilibrium of bodies which remain at rest under the action of forces have tendency to cause translation.<\/p>\n<p>\u00a0<strong>Condition of Equilibrium<br \/>\n<\/strong>When a block is placed on a table as shown below and force F1 and F2 are applied to the block.<br \/>\nThe block remains in translational equilibrium if the magnitude of F1 and F2 are equal.<\/p>\n<p>\u00a0Since F<sub>1<\/sub> and F<sub>2<\/sub> are acting in opposite direction, but have equal magnitudes.<br \/>\nThen, F<sub>1<\/sub> = -F<sub>2<\/sub><br \/>\n\t\t          F<sub>1<\/sub> + F<sub>2<\/sub> = 0<br \/>\nAlso, the upward force N balances the downward force mg on the block.<br \/>\n:.      N = -mg, N + mg  = 0<br \/>\nHence, the sum of the vertical components and horizontal components of forces acting on a body in translational equilibrium is equal to zero.<\/p>\n<p>\u00a0<strong>Example<br \/>\n<\/strong> 1. A particle of mass 5kg is supported by two light inelastic strings inclined at angles 30<sup>0<\/sup> and 45<sup>0<\/sup> respectively to the horizontal.  If the system is in equilibrium, calculate the tension in each string.<br \/>\nSolution:<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1859_Week5SS3Fi1.png\" alt=\"\"\/><\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0Let the two inelastic strings to be DP and OQ<br \/>\nResolve each string vertically and horizontally:<br \/>\nHorizontal : Efx  =  T<sub>1<\/sub>Cos 30<sup>0<\/sup> + T<sub>2<\/sub>cos 45<sup>0<\/sup> + 0<br \/>\nVertical    :   Efy  =   T<sub>1<\/sub> Sin 30<sup>0<\/sup>  + T<sub>2<\/sub> sin 45<sup>0<\/sup> \u2013 50<br \/>\n:.  Efx   = 0.7061T<sub>2<\/sub> \u2013 0.866T<sub>1<\/sub>   = 0   equation                  I<br \/>\nEfy   =  0.7071T<sub>2<\/sub>  + 0.5T<sub>1<\/sub>      = 50    equation              II<\/p>\n<p>\u00a0Solving equation I and II simultaneously:<\/p>\n<ul>\n<li>1.366T1  = -50\n<\/li>\n<\/ul>\n<p>    T<sub>1<\/sub>  = -50\/-1.366        T<sub>1<\/sub> = 36.6N<br \/>\nSubstitute T1 in equation I or II using equation 1, 0.7071T<sub>1<\/sub> \u2013 0.866T<sub>1<\/sub> = 0.<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a00.7071T<sub>2<\/sub> = 0.866 x 36.6<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0T<sub>2<\/sub> = 31.6956<br \/>\n                                                                                  0.7071<br \/>\n                                                                       T<sub>2<\/sub> = 44.82N<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0:. The tensions in the strings are T<sub>1<\/sub> = 36.6N, T<sub>2<\/sub> = 44.82N<\/p>\n<p>\u00a02.\u00a0\u00a0\u00a0\u00a0A particle of mass 5kg is suspended by a light in extensive string which makes an angle of 30<sup>0<\/sup> with the downward vertical and horizontal forced F.  If the system is in equilibrium, calculate <\/p>\n<ol>\n<li>The tension in the string       (ii) The magnitude of the force F (take g = 10ms<sup>-2<\/sup>)\n<\/li>\n<\/ol>\n<p>\u00a0<strong>Solution<\/strong><br \/>\n\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1859_Week5SS3Fi2.png\" alt=\"\"\/><br \/>\n\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1859_Week5SS3Fi3.png\" alt=\"\"\/><\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1859_Week5SS3Fi4.png\" alt=\"\"\/><\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0I using Sin \u03b8 = AB<br \/>\n\t\t                        OA<br \/>\nSin 60<sup>0<\/sup> = 5 x 10<br \/>\n                  T<br \/>\nT = 50\/sin 60<sup>0<\/sup>, T = 57.74N.   The tension in the string is 57.74 N<\/p>\n<p>\u00a0<br \/>\n\u00a0II. Magnitude of F<sub>1<\/sub> Tan 30<sup>0<\/sup> = OB<br \/>\n                                                 AB<br \/>\n         50 x Tan 30<sup>0<\/sup> = F\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0:.  F = 28.87N<\/p>\n<p>\u00a0<strong>EVALUATION<br \/>\n<\/strong>1.A street lamp of mass 10kg is suspended at a position ) by two wires OP and OQ across a rood such that each wire is inclined at an angle of 80<sup>0<\/sup> to the upward vertical,  If the system is in equilibrium, calculate the tension in one of the two wires.  (Take g = 10ms<sup>-2<\/sup>)<\/p>\n<p>\u00a0<strong>TRIANGLE OF FORCES<br \/>\n<\/strong>If three coplanar forces act on a body in such a way that the system is in equilibrium, then the forces can be represented in magnitude and direction by the sides of a triangle taken  in order<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1859_Week5SS3Fi5.png\" alt=\"\"\/><\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0The triangle representing the three coplanar forces is called a triangle of forces.<\/p>\n<p>\u00a0<strong>Example<\/strong>:<br \/>\n1.A body of mass 6.5kg is supported by two strings,  One of the stings is inclined at an angle of 30<sup>0<\/sup> and the other 40<sup>0<\/sup> to the horizontal .  Find the tension in each strings, if the system is in equilibrium (take g = 10ms<sup>-2<\/sup>)<\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<strong>Solution<\/strong><\/p>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1859_Week5SS3Fi6.png\" alt=\"\"\/><\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0Using Sine rule;<br \/>\nT1\u00a0\u00a0\u00a0\u00a0 = 65<br \/>\nSin 50<sup>o<\/sup>    Sin 70<sup>o<\/sup><br \/>\n\t\tT1       = \u00a0\u00a0\u00a0\u00a065 x sin 50<sup>o<\/sup><br \/>\n\t\t\t  Sin 50<sup>o<\/sup>             Sin 70<br \/>\nT<sub>1<\/sub> =   65 x 0.766<br \/>\n              0.9397<br \/>\nT<sub>1<\/sub> = 52.98N<br \/>\nSimilarly:<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1859_Week5SS3Fi7.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1859_Week5SS3Fi8.png\" alt=\"\"\/>T<sub>2<\/sub>      = \u00a0\u00a0\u00a0\u00a0W<br \/>\n\t\t\t Sin 60<sup>0<\/sup>\u00a0\u00a0\u00a0\u00a0Sin 70<sup>0<\/sup><br \/>\n\t\t T<sub>2<\/sub>, = 65 x sin 60<sup>0<\/sup>    = 65 x 0.8660<br \/>\nSin 70<sup>0<\/sup>                 0.9397<br \/>\n  T<sub>2<\/sub> = 59.9N<br \/>\n<strong>EVALUATION<em><br \/>\n\t\t\t\t<\/em><\/strong>A body of mass s10kg is suspended by means of two light inextensible strings. AP and BP which are inclined at angles 60<sup>0<\/sup> and 30<sup>0<\/sup> respectively to the downward vertical. If T<sub>1<\/sub> and T<sub>2<\/sub> are the magnitude of the tension AP and BP respectively, calculate the values of T<sub>1<\/sub> and T<sub>2<\/sub>.<\/p>\n<p>\u00a0<strong>LAMI&#8217;S THEOREM<\/strong><br \/>\n\t\tThis theorem states that if three forces acting at a point are in equilibrium, then each force is proportional to the sine of the angle between the liens of action of the other two forces.<br \/>\nConsider the forces F<sub>1<\/sub>, F<sub>2<\/sub> and F<sub>3<\/sub> below<\/p>\n<p>\u00a0By Lami&#8217;s theorem:  F<sub>1<\/sub> \u03b1 Sin <strong><em>B<\/em>F<sub>2<\/sub>  \u03b1<\/strong> sin <em>\u03b3<\/em>F<sub>3<\/sub>  \u03b1 sin \u03b8<strong><br \/>\n\t\t\t<\/strong>Since the forces are proportional to the sine of the angle: then   F<sub>1<\/sub>      =      F<sub>2<\/sub>  =     F<sub>3<\/sub><br \/>\n\t\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1859_Week5SS3Fi9.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1859_Week5SS3Fi10.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1859_Week5SS3Fi11.png\" alt=\"\"\/>                                                                                                   Sin <em>B<\/em>sin\u03b3        sin \u03b8<br \/>\nExample: <\/p>\n<ol>\n<li><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1859_Week5SS3Fi12.png\" alt=\"\"\/>A body of weight 91N is suspended by two inelastic string 5m and 12 m long attached to two points on the same horizontal level, whose distance apart is 13m.  Using Lami&#8217;s theorem or otherwise, find the tension along the strings.\n<\/li>\n<\/ol>\n<p><strong>Solution<\/strong><\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0s<\/p>\n<p>\u00a0Le the tension in the strings be T<sub>1<\/sub>nd T<sub>2<\/sub> and x, B be the angles made with the horizontal.<br \/>\n       AB<sup>2<\/sup>  = OA<sup>2<\/sup> + OB<sup>2<\/sup>  = 5<sup>2<\/sup> +12<sup>2<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0   =  25 + 144 =169<br \/>\n\u00a0\u00a0\u00a0\u00a0but AB = 13<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0AB<sup>2<\/sup>  = 13<sup>2<\/sup>  = 169<br \/>\n\u00a0\u00a0\u00a0\u00a0:. OA<sup>2<\/sup>  + OB<sup>2<\/sup>   = AB<sup>2<\/sup><br \/>\n\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1859_Week5SS3Fi13.png\" alt=\"\"\/>hence \u00a0\u00a0\u00a0\u00a0    AOB  is a right angle triangle<br \/>\n:.  Sin B = <sup>5<\/sup>\/<sub>13<\/sub>,  sin x  = <sup>12<\/sup>\/<sub>13<\/sub><br \/>\n\t\t  B = sin <sup>-1<\/sup> 0.3845\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u04e8 = sin<sup>-1<\/sup> 12\/13<br \/>\n  B = 22.6<sup>0<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0  = 0.9231  = 67.4<sup>0<\/sup><br \/>\n\t\tUsing Lami&#8217;s theorem:<br \/>\nT1\u00a0\u00a0\u00a0\u00a0      =\u00a0\u00a0\u00a0\u00a0W<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1859_Week5SS3Fi14.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1859_Week5SS3Fi15.png\" alt=\"\"\/>Sin 157.4\u00a0\u00a0\u00a0\u00a0sin 90<sup>0<\/sup><br \/>\n\t\tT<sub>1<\/sub>  = 91 x sin 157.4<sup>0<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 91 x 0.3843<br \/>\n                Sin 90<sup>0<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01<br \/>\nT<sub>1<\/sub>  = 34.97N<br \/>\nT<sub>2<\/sub>\u00a0\u00a0\u00a0\u00a0       =  \u00a0\u00a0\u00a0\u00a0 W<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1859_Week5SS3Fi16.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1859_Week5SS3Fi17.png\" alt=\"\"\/> Sin 112.6 \u00a0\u00a0\u00a0\u00a0   sin 90<sup>0<\/sup>          T<sub>2<\/sub>  = 91 x sin 112.6<br \/>\n                                                             Sin 90<sup>0<\/sup><br \/>\n\t\tT<sub>2<\/sub> =  91 x 0.9232<br \/>\n                1<br \/>\n T<sub>2<\/sub> = 84.01N.<\/p>\n<p>\u00a0<strong>EVALUATION<\/strong><br \/>\n\t\tA particle of mass 10kg is connected by two strings of length 3m and 4m to two points on the same horizontal level and 5m apart, find the tension in the strings.<\/p>\n<p>\u00a0<strong>GENERAL EVALUATION<br \/>\n<\/strong><\/p>\n<ol>\n<li>A particle of mass 98kg is suspended by two light inelastic strings of length 9m and 12m from two fixed point P and which are 15m apart.  Calculate (i) the angles made by the strings with the upward vertical (ii) the tension in the strings.\n<\/li>\n<li>The ends P and Q of an inextensible string 17m long are attached to two fixed points 13m apart on the same horizontal level. A body of mass 20kg is suspended from a point C on the string 5m from P. Calculate  (a) the angle which each part of the string makes with the horizontal  (b) the tension in each part of the string.\n<\/li>\n<\/ol>\n<p>\u00a0<strong>READING ASIGNMENT<br \/>\n<\/strong>Read Equilibrium &#8220;pages&#8221; 170-177 of Further Mathematics Project III.<\/p>\n<p>\u00a0<strong>WEEKEND ASSIGNMENT<\/strong><br \/>\n\t\tTwo forces (8N, 030<sup>0<\/sup>) and (10N, 120<sup>0<\/sup>) act on a body; find the magnitude of the force that would be applied to keep the system in equilibrium.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>\u00a0WEEK FIVE STATICS \u2013 CONTINUATION Definition of equilibrium Condition of equilibrium of rigid body Application&#8230;<\/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-3791","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\/3791","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=3791"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3791\/revisions"}],"predecessor-version":[{"id":3792,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3791\/revisions\/3792"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=3791"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=3791"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=3791"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}