{"id":3789,"date":"2023-10-05T18:58:29","date_gmt":"2023-10-05T18:58:29","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=3789"},"modified":"2023-10-05T19:03:00","modified_gmt":"2023-10-05T19:03:00","slug":"week-4-ss3-first-term-further-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-4-ss3-first-term-further-mathematics-notes\/","title":{"rendered":"Week 4 &#8211; SS3 First Term Further Mathematics  Notes"},"content":{"rendered":"<p>\u00a0<strong>WEEK FOUR<br \/>\n<\/strong><strong>STATICS<br \/>\n<\/strong><\/p>\n<ul>\n<li>Definition of Concepts\n<\/li>\n<li>Resultant of two forces\n<\/li>\n<li>Components resolution of forces\n<\/li>\n<\/ul>\n<p><strong>DEFINITION OF CONCEPT<br \/>\n<\/strong>Statics is the study of bodies which remain at rest under the action of given forces.<\/p>\n<p>\u00a0<strong>Mass:<\/strong> This is the quantity of matter contain in a body.  Mass of a particular body does not change and the standard unit is kilogram.<\/p>\n<p>\u00a0<strong>Force:<\/strong> Force is that action which tends to change the state of rest or uniform motion of a body in a straight line.  It&#8217;s a vector quantity sine it has magnitude and direction.  The unit of force is Newton.<br \/>\n:.  F = Ma<br \/>\nWhere M =  Mass,  a = acceleration.<br \/>\nComposition of Forces:  Two or more concurrent forces can be combined to obtain a single force.  Therefore, resultant force is the force produced or obtained when two or more concurrent forces are combined.<br \/>\nA force can be resolve by<br \/>\nI Graphical Method\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0II. Analytical Method.<\/p>\n<p>\u00a0<strong>Analytical Method<\/strong>:  The parallelogram law of composition of two forces is used to find the resultant force of two or more forces.  Hence, parallelogram law states that if two forces acting at a point are represented in magnitude and direction by two adjacent sides of a parallelogram, then the resultants of the two forces, is represented in magnitude and direction by the diagonal of the parallelogram, drawn from the point of action of the two forces.<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi1.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi2.png\" alt=\"\"\/>                                         A                                                C<\/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_1858_Week4SS3Fi3.png\" alt=\"\"\/><br \/>\n\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi4.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a0                           B                                                             D<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi5.png\" alt=\"\"\/><br \/>\n\t\tR is the resultant force and can be obtained using cosine rule:<br \/>\nR<sup>2<\/sup> = P<sup>2<\/sup> + Q<sup>2<\/sup> \u2013 2PQcos(180 \u2013\u03b8)<br \/>\nThe angle of inclination is \u03ac and can be obtained using sine rule or tan.<br \/>\n:. Tan \u03ac = CD<br \/>\n                OD<br \/>\nTan \u03ac = P Sin \u03b8<br \/>\n             Q + P Cos \u03b8<\/p>\n<p>\u00a0<strong>Example 1<\/strong>:  The angle between two forces of magnitude 8N and 5N is 120<sup>0<\/sup>. Find in N, the magnitude of their resultant.<br \/>\nSolution <\/p>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi6.png\" alt=\"\"\/>                          P                                                                R<\/p>\n<p>\u00a0<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0 5N\u00a0\u00a0\u00a0\u00a0R<\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0                     Q\u00a0\u00a0\u00a0\u00a08N                           S<br \/>\nLet R be the resultant vectorial force.<br \/>\nR<sup>2 <\/sup>= P<sup>2<\/sup> + Q<sup>2<\/sup> \u2013 2PQ Cos R.<br \/>\n     = 5<sup>2<\/sup> + 8<sup>2<\/sup> \u2013 2 (5 x 8) Cos 60<sup>o<\/sup><br \/>\n\t\t      = 25 + 64 \u2013 80 x \u00bd<br \/>\n      = 89 \u2013 40<br \/>\n                   R<sup>2<\/sup> = 49<br \/>\n:. R = \u221a49               R = 7N<br \/>\n2. Calculate, correct to one decimal place, the angle between two forces 20N and 30N if their resultant is 40N.<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi7.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi8.png\" alt=\"\"\/><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi9.png\" alt=\"\"\/><br \/>\n\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi10.png\" alt=\"\"\/><\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0R<sup>2<\/sup> = P<sup>2<\/sup> + Q<sup>2<\/sup> \u2013 2PQ Cos (180 \u2013 \u03b8)<br \/>\n\u00a0\u00a0\u00a0\u00a040<sup>2<\/sup> = 20<sup>2<\/sup> + 30<sup>2 <\/sup>\u2013 2 (20 x 30) Cos (180 \u2013 \u03b8)<br \/>\n\u00a0\u00a0\u00a0\u00a01600 = 400 + 900 \u2013 1200 cos(180 \u2013 \u03b8)<br \/>\n\u00a0\u00a0\u00a0\u00a01600-1300 = 1200Cos (180 \u2013\u03b8)<br \/>\n\u00a0\u00a0\u00a0\u00a0300 = -1200 Cos (180 \u2013 \u03b8)<\/p>\n<ol>\n<li>=  Cos ( 180 \u2013\u03b8 )\n<\/li>\n<\/ol>\n<p>           -1200<br \/>\n            -1\/4   = Cos (180 \u2013 R)<br \/>\n            &#8211; 180 \u2013\u03b8 = Cos <sup>-1<\/sup>(<sup>-1<\/sup>\/<sub>4<\/sub>)<br \/>\n             180 \u2013 \u03b8 = 104. 5<br \/>\n              180 \u2013 104.5   = \u03b8<br \/>\n               \u03b8 = 75.5<br \/>\n    :. The angle between them is 75.5<sup>o<\/sup><\/p>\n<p>\u00a0<strong>EVALUATION<br \/>\n<\/strong>The angle between two forces of magnitude 8N and 11N is 35<sup>o<\/sup>. Find the magnitude and inclination to the 11N force of resultant force.<\/p>\n<p>\u00a0<strong>RESOLUTION OF FORCES<br \/>\n<\/strong>A given force can be resolved into two parts and each part is called the resolute.  A given force can be resolve in two directions which are perpendicular to each other.<br \/>\nThe component along the y axis is called the vertical component and the component along the x- axis is the horizontal component.<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi11.png\" alt=\"\"\/><br \/>\n\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi12.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi13.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi14.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0y<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi15.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi16.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi17.png\" alt=\"\"\/><br \/>\n\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi18.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi19.png\" alt=\"\"\/><br \/>\n\t\tpyPy<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi20.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi21.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0\u03b8\u00a0\u00a0\u00a0\u00a0<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi22.png\" alt=\"\"\/><br \/>\n\t\tPxPx<br \/>\nLet Px the horizontal component<br \/>\nLet py the vertical component.<br \/>\nIf p is inclined to the upward vertical<br \/>\n<img decoding=\"async\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi23.png\" alt=\"\"\/><br \/>\n\t\tPx Sin \u03b8 = pxCos \u03b8 = py<br \/>\n  PP<br \/>\nPx  = P sin \u03b8.                                 Py = P cos \u04e8.<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi24.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi25.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi26.png\" alt=\"\"\/>Py                                    P<sub>1<\/sub><\/p>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi27.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0P4<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi28.png\" alt=\"\"\/><br \/>\n\t\tpy<\/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_1858_Week4SS3Fi29.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi30.png\" alt=\"\"\/><br \/>\n\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi31.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi32.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi33.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0px<br \/>\n\u00a0\u00a0\u00a0\u00a0-px\u00a0\u00a0\u00a0\u00a0px<\/p>\n<p>\u00a0<br \/>\n\u00a0                                                     P<sub>3<\/sub>\u00a0\u00a0\u00a0\u00a0Py             P<sub>2<br \/>\n<\/sub><br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0                         Force\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Horizontal Component\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Vertical Component<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi34.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi35.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi36.png\" alt=\"\"\/><br \/>\n\t\tP1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0P1 Sin \u00a0\u00a0\u00a0\u00a0\u03b8\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0  P1 Cos \u03b8<br \/>\nP2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0P2cos \u03b8\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-P2 Sin\u03b8<br \/>\nP3\u00a0\u00a0\u00a0\u00a0           -P3 cos \u03b8\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0&#8211; P3 Sin \u03b8<br \/>\nP4\u00a0\u00a0\u00a0\u00a0          &#8211; P4 cos \u03b8\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0  P4 sin \u03b8<br \/>\n\u00a0\u00a0\u00a0\u00a0<br \/>\n\u00a0Resultant of several concurrent forces :the resultant is obtained as :<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi37.png\" alt=\"\"\/><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0R = \u221a(\u2211P<sub>x<\/sub>)<sup>2<\/sup>  + (\u2211P<sub>y<\/sub>)<sup>2<br \/>\n<\/sup>Where \u2211Px = Sum of horizontal components       \u2211Py  = Sum of vertical components<br \/>\nTan \u03b8   =Py<br \/>\nPx<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi38.png\" alt=\"\"\/><br \/>\n\t\t  \u0398 = tan <sup>-1<\/sup>   \u2211Py<br \/>\n\t\t\u2211Px\u00a0\u00a0\u00a0\u00a0Angle of inclination to the resultant.<\/p>\n<p>\u00a0<strong>Example<\/strong>:<br \/>\n1.The horizontal component of a force p which makes an angle of 50o with the horizontal is 30N. Find the force P.<\/p>\n<p>\u00a0<strong>Solution<em><br \/>\n\t\t\t\t<\/em><\/strong>30 = P x 0.6 + 28<br \/>\n    P = 30\/0.6428<br \/>\n    P = 46.67N<br \/>\n2.Three forces (2N, 060<sup>0<\/sup>), ( 4.5N, 180<sup>0<\/sup>) and (5N, 300<sup>0<\/sup>) act on a body of mass 2kg which is initially at rest. Find I the resultant force on the body II the acceleration with which the body begins to move.<br \/>\n<strong>Solution<\/strong><br \/>\n\t\tLet P1 = (2N.060<sup>0<\/sup>).  P2 = (4.5N, 180<sup>0<\/sup>)   P3 = (5N, 300<sup>0<\/sup>)<\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi39.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi40.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi41.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi42.png\" alt=\"\"\/><\/p>\n<p>\u00a0                               5N                                               2N<\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0                                                            4.5N<\/p>\n<p>\u00a0<br \/>\n\u00a0       Horizontal Component\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Vertical Components<br \/>\n       P1  = 2 Sin 60<sup>0<\/sup> = 1.7321\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0P1 = 2 Cos 60<sup>0<\/sup> = 1<br \/>\n       P2  = 4.5 Cos 90<sup>0<\/sup> = O\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0P2 = -4.5 Sin 90<sup>0<\/sup> = -4.5<br \/>\n       P3  = 5Cos 30<sup>0<\/sup> = -4.33\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0P3 = 5 Sin 30<sup>0<\/sup> = 2.5<br \/>\n      \u2211px  = 1.732 + 0 \u2013 4.33\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0          \u2211py = 1 \u2013 4.5 + 2.5<br \/>\n\u00a0\u00a0\u00a0\u00a0  = &#8211; 2.598\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0  \u2211py  = -1<br \/>\n\u00a0\u00a0\u00a0\u00a0 (\u2211Px)<sup>2<\/sup> = 6.7496\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0   (\u2211py)<sup>2<\/sup> = 1<\/p>\n<p>\u00a0:. R  =\u221a (\u2211Px )<sup>2<\/sup>+ (\u2211Py)<sup>2<\/sup>= \u221a6.7496 + 1    =\u00a0\u00a0\u00a0\u00a0\u221a7.7496<br \/>\n  R  = 2.78N<\/p>\n<p>\u00a0<\/p>\n<ol>\n<li>\n<div>Acceleration of the body ; F = ma\n<\/div>\n<p>M = 2kg\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0F = Resultant force = 2.78<br \/>\nF = ma\n<\/li>\n<\/ol>\n<ol>\n<li>\n<div>            = 2 x a\n<\/div>\n<p>a = <sup>2.78\/<\/sup><sub>2\u00a0\u00a0\u00a0\u00a0<\/sub>     a  = 1.39ms<sup>-2<\/sup><br \/>\n\t\t\t\t<strong>EVALUATION<\/strong><br \/>\n\t\t\t\tThe vertical component of a force F which makes an angle of 35<sup>0<\/sup> with the horizontal is 45N.  Find the force F.<\/p>\n<p>\u00a0<strong>GENERAL EVALUATION<br \/>\n<\/strong><\/p>\n<ol>\n<li>\n<div>The forces of magnitude 35N and 45N act on a particle in the directions 180<sup>0<\/sup> and 315<sup>0<\/sup> respectively.  Find the resultant of these forces giving:\n<\/div>\n<p>a.   the magnitude correct to the nearest whole number   b. the direction correct to the nearest degree.\n<\/li>\n<li>A vertical force of 6N and a horizontal force of 8N act on a body.  Find the magnitude and the inclination of the resultant force to the horizontal.\n<\/li>\n<\/ol>\n<p>\u00a0<strong>READING ASIGNMENT<br \/>\n<\/strong>Read Composition and Resolution of Coplanar forces on pages 154 to 165 of further mathematics project III.<\/p>\n<p>\u00a0<strong>WEEKEND ASSIGNMENT<br \/>\n<\/strong>1. \u00a0\u00a0\u00a0\u00a0Two forces each of magnitude PN are inclined to each other at an angle of 120<sup>0<\/sup>. Find the magnitude of their resultant.<br \/>\n\u00a0\u00a0\u00a0\u00a0(a) P\u221a3 N\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) P2N\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0( c) PN<br \/>\n2.\u00a0\u00a0\u00a0\u00a0Find the angle between the two forces 5N and 6N if their resultant is 8N.<br \/>\n\u00a0\u00a0\u00a0\u00a0(a) 60<sup>0<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) 120<sup>0<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(c) 180<sup>0<\/sup><br \/>\n\t\t\t\t3.\u00a0\u00a0\u00a0\u00a0A force P of magnitude 60N makes an angle of 40<sup>0<\/sup> with the horizontal. Use the information to answer questions 3 and 4<\/p>\n<p>\u00a03.\u00a0\u00a0\u00a0\u00a0Find the horizontal component of P<br \/>\n\u00a0\u00a0\u00a0\u00a0(a) 20N\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) 45.96N\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(c ) 38.57N<br \/>\n4.\u00a0\u00a0\u00a0\u00a0Find the vertical component of P<br \/>\n\u00a0\u00a0\u00a0\u00a0(a) 45.96N\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) 38.57\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(d) 20N.<br \/>\n5.\u00a0\u00a0\u00a0\u00a0Find the resultant of forces 8N and 10N inclined at an angle 120<sup>0<\/sup> to each other.<br \/>\n\u00a0\u00a0\u00a0\u00a0(a) 2\u221a61N\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) 61\u221a2N\u00a0\u00a0\u00a0\u00a0  C 39N<\/p>\n<p>\u00a0<strong>THEORY<br \/>\n<\/strong><\/p>\n<ol>\n<li>Forces F1 = (10N, 090<sup>0<\/sup>), F2 = (20N, 210<sup>0<\/sup>) and F3 = (4N, 330<sup>0<\/sup>) act on a body  at rest on a smooth table.  Find, correct to one decimal place the magnitude of the resultant force.\n<\/li>\n<li>Find the magnitude and direction of the resultant  of the forces shown in the diagram below:\n<\/li>\n<\/ol>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi43.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi44.png\" alt=\"\"\/><br \/>\n\t\t\t\t                                                   6N                             8N<\/p>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100523_1858_Week4SS3Fi45.png\" alt=\"\"\/><\/p>\n<p>\u00a0<br \/>\n\u00a0                                                             10N<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>\u00a0WEEK FOUR STATICS Definition of Concepts Resultant of two forces Components resolution of forces DEFINITION&#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-3789","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\/3789","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=3789"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3789\/revisions"}],"predecessor-version":[{"id":3790,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3789\/revisions\/3790"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=3789"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=3789"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=3789"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}