Maths: Pre-medicine and Pre-pharmacy PDF

Dr. Mohamed Abdulai Koroma Senior Lecturer of Mathematics Fourah Bay College University of Sierra Leone. Pre-medicine 1 and Pre-pharmacy Mathematics 1. Plane Geometry 1.1 angles Definition 1.1 an angle is said to be acute when it lies between 0 and 90 degrees. i.e 0  ...

Dr. Mohamed Abdulai Koroma Senior Lecturer of Mathematics Fourah Bay College University of Sierra Leone. Pre-medicine 1 and Pre-pharmacy Mathematics 1. Plane Geometry 1.1 angles Definition 1.1 an angle is said to be acute when it lies between 0 and 90 degrees. i.e 0    90. Definition 1.2 an angle is said to be obtuse when it lies between 90 and 180 degrees. i.e 0    180 Definition 1.3 an angle is said to be reflex when it is more than 180 degrees. i.e 180     (i) (ii)   (iii) 1. Units of Measurement of Angles Angles are measured either in degrees 0 or in radians  c 1.3 Angles on a Straight Line Angles on a Straight Line sum up to 180 c  c  c  c       180 1.3 Parallel Line Any straight lines on the same plane which era equidistant from each other (do not intersect) are parallel. 1.4 Intercept Theorems When two parallel lines intersect a  third line,(1) alternate angles are equal; (2) corresponding angles are equal.  (1)   alternate angles     (2)  Corresponding angles  When two lines intersect, vertically opposite angles are equal     Vertically opposite angles Draw line PAQ parallel to CB. QAB  ABC (alternate angles) PAC  ACB (alternate angles) Now       180 (angles in a straight line) And this completes the proof. (2) Exterior angle in a triangle equals the sum of interior opposite angles. Proof : A D C B 1.5 Triangles (1) The sum of the angles in a triangle is equal to 180 degrees. A  c   P B  Proof: A  Q    C B DCA ACB  180.........(1) ( angleson a straightline) CAB ABC ACB  180.....(2)(anglesin a triangle) from (1) and (2) we have DCA ACB  CAB ABC ACB  DCA  CAB ABC And this completes the proof. 1.6 Properties of Special triangles 1.6.1 In an equilateral triangle all the angles are equal and all the sides are of the same length.  A 600 600 60 0 C B 1.6.2 In an equilateral triangle the bisector of each angle bisects the opposite side and is perpendicular to it. 30 0 30 0 1.6.3 In an isosceles triangle two sides are equal and their base angles are equal. A ACB  ABC and AC  AB C B 1.6.4 In an isosceles triangle the bisector of each of the base angles intersects the opposite side at right angles and divides it into two equal halves. A C B 1.6.5 In a right angled triangle, one of the three angles is equal to 90 degrees and the side facing the right angle is called the hypotenuse. Hypotenuse Right angle 1.6.7 In a right angled triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides (Pythagoras' theorem) A AC  AB  AC 2 2 2 B C 1.6.8 In any triangle the length of one side is not grater than the sum of the lengths of the other two sides. That is, if ABC is a triangle AB  AC  BC. Exercise 1.0 solve for the unknowns in each of the following: 1. 30 0  0 140 x 70 0 2. y 0 z a 150 3. 600 x y 920 z w 4. 920 150 5. 70 x y  60 6. z 3x 120 w 60 x 7. 25 25 a 85 y 30 z x 8 80 y a 1.7 Similar Triangles Two triangles are said to be similar when: (i) their angles correspond and are equal in pairs; (ii) the lengths of the sides of one triangle are proportional to those of another. Consider the following triangles: A D F C B E CAB  FDE; ABC  DEF ; BCA  EFD The sides are proportional and so AC AB BC   DF DE EF If two triangles are similar the ratio of their areas is equal to the square of the ratio of corresponding sides. i.e. area of ABC  AB  AC  BC 2 2 2 area of DEF DE DF EF Example1.7.1 consider the figure Below. A E B D C Given that EB//DC show that ABE  ACD. Proof: since EB//DC  AEB   ADC (correspon ding angles)  ABE   ACD (correspon ding angles)  EAB   DAC (common) Thus ACD  ABE Exercise 1.7.1 (i)consider the figure below. A E D C B Given that CED  ABC , show that ABD  ECD (ii)consider the figure below.  A find DE, CE and BD 10cm 8cm B 7cm C D E Exercise 2.7.2 Find the unknown in each of the following: (i) A ycm 10 cm B zcm 5 cm E xcm C D 8 cm (ii) z (iii) x x 8 cm z0 0 140 12 cm ycm y 0 2. Properties of Four sided figures 1. A rectangle is a four sided figure in which: (i) opposite sides are parallel; (ii) opposite sides are parallel and equal in length; (iii) adjacent sides intersect at right angles; (iv) the diagonals bisect each other and intersect at the centre. 2. A square has all the futures of a rectangle. Here, all the sides are of equal length and the diagonals bisect at the centre at right angles. 3. A parallelogram is a four sided figure in which: (i) opposite side are parallel and equal in length; (ii) the diagonals bisect at the centre and are equal in length. 4. In Rhombus all the sides are of equal length and the diagonals bisect each other at right angles. 5. In a trapezium two of the sides are parallel. Exercise 2.7.3 Solve for the unknowns in each of the following: 1. (a) 110 0 c a b e (b) z 0 (c) z0 30 0 y0 1200 x0 730 1600 x0 (d) z 0 y0 (e) y0 x0 1300 65 0 26 0 (f) (g) 0 y0 z x0 0 y w0 v0 x0 0 0 53 u0 w0 z 24 0 2.a 8cm xcm zcm 3cm b) 10cm ycm ycm  0 7cm 12cm xcm 8cm in (b) the area of the rectangle is twice the area of the triangle. (c) (d) 12cm 18cm 6cm xcm 7.2cm 32cm xcm ycm 15.9cm 7.6cm zcm 2.4cm 3.4cm 3. Perimeters and Areas 3.1 The perimeter of a triangle is the sum of the lengths of its sides. c b a Perimeter p  a  b  c 3.2 The perimeter of a rectangle, square, trapezium, rhombus or parallelogram equals the sum of the lengths of its sides.  b d c a a a b P  abcd P  2(a  b) P  4a a b a P  4a P  2(a  b) 3.3 Area of a triangle (a) For a right angled triangle the area is half the product of the base and the height. h Area  1 b h 2 b (b) For a triangle with two given sides the area is half the product of the lengths the two sides and the sine of the given angle. 1 a  b Area   a  b sin  2 (c) For a triangle in which the lengths of all the three sides are known we find the semi perimeter and determine the area as given below:  1 b s  a  b  c   semi perimeter c 2 a Area  s ( s  a )( s  b)( s  c) 3.4 The area of a rectangle is equal to the product of the length and the width. l b Area  l  b 3.5 The area of a square is equal to the product of the square of one side.  Area  l 2 l 3.6 Area of a trapezium equals half the sum of the lengths of the parallel sides multiplied by the height. b 1 Area  (a  b)  h h 2 a 3.7 Area of a parallelogram equals the product of the base and the height. a h Area  b  h, where h  a sin  b 3.8 The area of a rhombus equals half the product of the lengths of its diagonals. b a 1 Area   a  b 2 Exercise 3.1 find the area and the perimeter of each of the following figures: 1. 5cm 4.6cm 2. 6.6cm 7.2cm 6.7cm 3. 10.7cm 4. 60 0 450 300 28cm 45cm 5. 36.3cm 30.8cm 37.2cm 67.2cm 26cm 18.4cm 4. Circle theorems 4.1 The angle between a tangent and the radius to a circle is 900. O T A 4.2 Angles subtended by an arc or a chord of a circle on the same segment D are equal. A C B 4.3 Angle in a semi circle is equal to 90 0 A C B 4.4 Opposite angles in a cyclic quadrilateral are supplementary(sum up to 1800) A B DAB  BCD  180 0 ABC  CDA  180 0 D C 4.5 Angle between a chord of a circle and a tangent is equal to the angle in the alternate segment. A B DCB  BAC D C 4.6 Angles subtended on the circumference of a circle by chords or arcs of equal lengths are equal.  F A AB  ED E and ACB  FDE D B C 4.7 The angles subtended by chords or arcs of equal lengths at the centre of a circle are equal. A o D AB  DC and AOB  DOC B C 4.8 The angle subtended by a chord or an arc at the centre of a circle is twice that on the alternate A segment. O C B Exercise 4.1 Find the value of the unknown in each of the following:  180 a0 0 0 z0 x z b0 0 y0 x0 y 0 38 1200 Show that x  y  z  90 0 w0 z0 y0 y0 x0 x 0 1300 27 0 820 z0 APPLIED MAHEMATICS 1. DYNAMICS KINEMATICS OF A PARTICLE MOVING IN A STRAIGHT LINE Consider a particle projected with speedu so that it covers a distance s after time t reaching a speed of v and an acceleration of a. (i) The average speed of the particle is given by  sum of speeds total distance average   2 time uv s  2 t  t (u  v)  2 s.............(1) (ii) The acceleration of the particle is velocity change given by acceleration  time v u a t   at  v  u  v  u  at...............(2) Using (2) on the left hand side of (1) gives t (u  u  at )  2 s  t (2u  at )  2 s 1 2  s  ut  at.........(3) 2 Now vu vu a t ............(4) t a Substituting (4) in (1) gives  v u   u  v   2 s  a  v  u  2as 2 2  v  u  2as..............(5) 2 2 (2), (3) and (5) are the equations of Motion for a particle moving with constant acceleration along a straight path. Example 1. A particle is moving on a straight path has an initial speed of u and an acceleration a. The particle covers a distance s after time t. (i) Given that given that v  7 , u  5 , a  1, find s and t (ii) If u  10 , v   2 , t  4 , find s and a (iii) Given that u  50, s  300, a  4, find t and v (iv) if v  60, s  45, a  40, find t and u (v) If v  18, s  64, t  8, find u and a Example 2. a car starts from rest and accelerates uniformly to a speed of 60km / h in 30 seconds. Find the distance covered by the particle in this time. Solution to example 1. (i) v  7, u  5, a  1, to find s and t We consider the equation v  u  2as 2 2  7  5  2(1) s 2 2 49  25  2 s 49  25 24 s   12 2 2 From v  u  at We have 7  5  (1)t  t  75  2 (ii) u  10 , v   2 , t  4 , to find s and a we consider the equation v  u  at  2  10  a (4)  4a  2  10  12 a  3 4 Solution to example 2. Since the car starts from rest u  0. v  60km/h, t  30 sec. To find s we consider the equation v  u  at  60  0  30a 60 a 2 30 1 2 We now consider s  ut  at 2 1  s  0(30)  (2)(30) 2 2 s  900km Example 3. A particle is projected upward from a point O with speed 25ms 1 Find the maximum height reached by the particle and the time that elapsed the time that elapsed before it returns to O. Solution: u  20, s  h, For the upward motion a   g  10 At maximum height v  0 and thus using the equation v  u  2as 2 2 We have 0 2  252  2 gh  0  625  2(10)h  10h  625  h  62.5m The time taken to reach maximum height is given by v  u  at  0  25  ( g )t  0  25  10t  10t  25 t  2.5 The particle takes the same time to reach maximum height and to return to the point o projection from maximum height. Thus the total time of flight is T  2  2.5  5s Example 4. A boy on a bridge throws a stone vertically upwards with a speed of 6m / s. 2 seconds later the stone hits the water below. Find the speed with which the particle hit the water and the initial height of the stone above the water. H  maximum height u  6, a   g  10, h  initial heightabove water t  2, s   h, v  ? We treat this as the case of a particle moving up continuously for 2 seconds. The initial distance of the stone above the water is negative. 1 2 From s  ut  at 2 1 We have h  6(2)  (10)(2 ) 2 2 h  12  20 h  8m From v 2  u 2  2as We have v  6  2(10)(8) 2 2 v  36  160 2 v   196  14 Thus, the stone hits the water with a speed of 14m / s and the initial height of the stone above the water is 8m Exercise 1. The points P, Q, R and S lie on a straight line such that QR=28m and RS=72m. A Particle starting from rest at P moves with constant acceleration and passes through Q, R and S. Its speeds at R and S are 9ms and 15ms respectively. Find 1 1 the speed of the particle at Q and determine how long it takes to travel from Q to S. Exercise 2. (a) a particle accelerates from rest at a constant rate of 3ms 1 to a speed of Vms. It mentioned this 1 speed for some time and then decelerate at a constant rate of 1.5ms.2 If the total time taken is one minute and the total distance travelled is one km find the value of V. (b) Two stations P and Q are 10km apart a car travels at a constant speed of 144km / h passes station P at 9 : 00 hours. At a distance skm from P the car applied its brakes, producing a  2 constant retardation of 0.4ms. Given that the car comes to rest at P, find s and determine when the car reaches P. (c) A particle which is moving in a straight line with constant acceleration covers a distance of 10m in in 2s , it then traveled a further distance of 20m in the next 2s. Find the further distance travelled in 3 more seconds and the final speed of the particle at the end of these 7 s. (d) A bus being brought to rest with a constant retardation travels 60m in 4s and a further 60m in 8s. Find the retardation of the train and the further time it takes before coming to rest. (e) A particle is projected vertically upwards from a point P rises 20m in the third second during its motion. Find the initial speed of the particle and the Maximum height reached by the particle from P. (f) A train sets off from a station P and accelerate uniformly for t1s and arrives at a point Q which is 300m. The train t1 then traveled at a speed of Vkm / h for t 2 s , covering a further distance of 1250m. It then decelerate uniformly for t3 s , coming to rest at another station R. If the total time taken is 3 min utes and 2t1  3t3 find the distance from P to R and the values of t1 , t 2 , t3 and V. (f) A stone is dropped from the top of a tower. After one second another stone is thrown vertically downwards from the same point at a speed of15m / s. If the stones reach the ground simultaneously , find the height of the tower. 2. Motion With Variable Acceleration In some motion, the distance, speed and acceleration may all be functions of time. In such situations we use the methods of calculus. Is s represents the distance, v the speed, a the acceleration and t the time have. ds (i) speed(velocity)= v ............(1) 2 dt ds d s (ii) acceleration a  2......(2) dt dt  ds (iii) v dt  vdt  ds   vdt   ds  s   vdt......................................(3) dv (iv) a dt  adt  dv   adt   dv  v   adt.......................................(4) Example 1. A particle p moves along a path such that its distance m from a fixed point O is given by the expression s  3t  14t  16t  10. Where t is the 3 2 time in seconds. Find (i) the distance of the particle from O after 2 seconds; (ii) the distance covered by the particle during the third second, (iii) the speed of the particle after 4 seconds, (iv) the instance when the particle is at rest and its distanse from O, (v) the acceleration of the particle after 3 seconds. Solution : (i) After 2 seconds t  2 ans so s  3t 3  14t 2  16t  10  s  3(2)  14(2)  16(2)  10  24  56  32  10 3 2 s  10m (ii) the distance covered during the third second is the Example 1. A particle p moves along a path such that its distance m from a fixed point O is given by the expression s  3t  14t  16t  10. Where t is the time 3 2 in seconds. Find (i) the distance of the particle from O after 2 seconds; (ii) the distance covered by the particle during the third second, (iii) the speed of the particle after 4 seconds, (iv) the instance when the particle is at rest and its distanse from O, (v) the acceleration of the particle after 3 seconds. Solution : (i) After 2 seconds t  2 ans so s  3t  14t  16t  10 3 2  s  3(2)  14(2)  16(2)  10 3 2  24  56  32  10  10m (ii) the distance covered during the third second is the distance between t  2 and t  3. At t  3, s  3(3)  14(3)  16(3)  10 3 2  81  126  48  10  13m Thus the distance covered during the third second is s  s3  s2  13  10  3m (iii) speed  v  ds dt  9t  28t  16 2 v  94   28(4)  16  144  112  16  48m / s 2 (iv) At rest the velocity is zero  9t  28t  16  0 2 28  (28)  4(9)(16) 2 so t  18 t  2.3568s or t  0.7543s When t  2.3568, s  3(2.3568)  14(2.3568)  16(2.3568)  10 3 2 s  9.2983  9.3m When t  0.7543, s  3(0.7543)  14(0.7543)  16(0.7543)  10 3 2 s  15.3908  15.4m  dv (v) a   18t  28 dt At t  3, a  18(3)  28  26m / s 2 Example2. The acceleration of a car whichis movingalongcurvedroadis a  12t  34 m/s. Given thatthe car is 2 movingwith a speedof 40 m / s when it is at a distanceof 10m froma gas stationat t  0, find : (i) the speed of the car at tiimet; (ii) the distance of the car from the gas station at any time t; (iii) the time when thecar is momentarily at rest and its distancefrom the gas station at this instance;(iv) the maximumspeed of the car and its distancefrom the gas station at this instane. Solution: (i) the speed of the car at timet is v   adt v   (12t  34)dt  6t  34t  c...................(1) 2 But v  40 at t  0 so (1) gives 40  6(0)  34(0)  c  c  40 2 Thus v  6t  34t  40 2 (ii) The distance from the gas satation at any time is   s   vdt   6t  34t  40 dt 2t  17t  40t  k 2 3 But s  10 at t  10 so k  10 Thus s  2t  17t  40t  10 3 (iii) When the car is at rest v  0 so we have 6t  34t  40  0  3t  17t  20  0 2 2  (t  4)(3t  5)  0 5 Therefore t  4s or t  s 3 When t  4 the distance is s  2(4)  17(4)  40(4)  10 3  128  68  160  60  230m  5 When t  the distanceis 3 3 2 5 5 5 s  2   17   40   10  3  3  3  38.7037m (v) the maximumspeed of th car occur when the acceleration is zero. 17 That is 12t  34  0  t  6 The distance of the car from the gas station is then 3 2  17   17   17  s  2   17   40   10 6 6 6  32.63519m Exercise 1. 1. Aparticle moving in a straight line such that its This image cannot currently be display ed. displacement from a fixed point after time t seconds is s  2t  13t  20t. Find the total distance 3 2 travelled and the maximum speed attained by the particle in the first 4 seconds of its motion. Find also the range of values of t for which the acceleration of the particle is negative. 2. A particle moving in a straight line starts from rest at a point A. Its acceleration at time t is 45  12t  9t m / s. If the particle comes to 2 2 rest instantaniously at a point B, find the distance AB and the time the particle takes to reach B. Find also the maximum acceleration and the maximum speed attained in this time. 3. Two cars P and Q move along a straight road fro a gas station G such that at time t their distance from the gas station are xP and xQ , respectively. If xP  5t (t  7) and xQ  t (t  3), 2 3 find (a) the time the cars meet again, (b) an expression for the velocity of P relative to Q, (c) the range of values of t for which P is moving faster than Q. 3.Due to track repairs a train retards 2 uniformly at 1m / s , from a speed of 40m / s at A to a speed of 10m / s at B. The train travels a distance of 3.5km, at a constant speed of 10m / s and then accelerates uniformly at 0.2m / s so that its speed at D is 40m / s. Sketch the velocity time graph for the journey from A to D, and show that the distance between these points is 8km. Show that the journey from A to D takes 330 s more than it would if the train travelled at a constant speed of 40m / s from A to D. 4. A particle moves along a curve such that its displacement from a fixed point O 1 3 7 2 is x  t  t  12t. Find : (i) the 3 2 distanceof the particle from O after 3 seconds; (ii) the speed of the particle at any value of t; (iii) when the particle is at rest and its distance from O; (iv) the maximum and the maximum acceleration of the particle. 2. Momentum and Kinetic energy (i) The momentum of a of a body is the product of ist mas and speed. momentum  mass  speed  mv The unit of momentum is the kgm / s (ii ) The kinetic energy is the energy possessed by an object by virtue of its motion. It is half the product of the mass and the square of the speed.  1 ke  mv  2 2 (iii) Principle of Linear Momentum : This states thet in a conservative system the total momentum is always the same. (iv) Momentum and Collision (direct impact) Suppose two particles of masses m1 and m2 moving initially with speeds u1 and u 2 collide directly so that their speeds after collision are v1 and v2. before impact after impact u1 u2 v1 v2 By the conservation of linear momentum we have m1u1  m2u 2  m1v1  v2u2 The change in kinetic energy due to impact is 1 2 1 2 1 1 2 m1v1  m2v2   m1u1  m2u2  2 2 2 2 2  1  m1v1  m2v2  m1u1  m2u2 2 2 2 2 2 Example 1. Two particles A and B of masses 2kg and 3kg moving in the same directions with speeds 8m / s and10m/ scollide directly. Given that the speed of B after impact is 6m/ s find the speed of A after impact. Find also the change in kinetic energy due to collision. Solution: consider the figures given below: Let the speed of A after impact be v1 before impact after impact 8 10 v1 6  ve A B A B By the law of conservation of linear momentum we have: m 1 u 1  m 2 u 2  m 1 v1  v 2 u 2  2 (8 )  3 (10 )  2 v1  3 ( 6 )  16  30  2 v1  18  2 v1  46  18  28  v1  14 total ke before impact  1 2 2 1 2 2 1 2  2 2  m1u1  m 2 u 2  2  8  3  10  214 total ke after impact  1 2 1 1   m1v1  m 2 v 2  2  14 2  3  6 2  250 2 2 2 2 loss in ke  ke after impact - ke before impact  250 - 214  36 J 2. Aparticle of mass 4 kg moving in a given direction with a speed of 5 m / s collides directly with a second particle of mass 7kg which is moving in the opposite direction with a speed of 10 m / s. Given that the particles stick toge ther after impact, find their common speed and the direction in which t he particles are are now movingm. solution : before impact after impact 5 10 v positive 4kg 7kg 11kg direction Let the positivedirectionbe as given above By thelaw of conservation of linear momentumwe have m1u1  m2u2  (m1  m2 )v  4(5)  7(10)  (4  7)v  20  70  11v  50  11v 50  v    4.545 11 The negative sign her indicates that the motion is in the reverse directio to that chosen Exercise : 1. A railway truck of mass 1200kg is moving along a straight level track at 7 m / s when it collides with another truck of mass 8000kg moving in the same direction at 2m / s. After the collision the trucks move on together. Find the common speed of the trucks after impact. 2.A bullet of mass 0.04kg is fired in to a block of wood of mass 1.4kg which rests on a horizontal surface. The bullet becomes embedded in the block And after impact they move on together at a speed of 20m/s. find the speed at which the bullet is fired. 3. Two particles, A of mass 4kg and B of mass 2kg, are moving horizontally along the same straight line when they collide. Both come to rest instantaneously. If the velocity of A just before impact was 3m/s, find the velocity of B and the loss in kinetic energy in the impact. 4. A gun of mass of mass 4kg fires a bullet of mass 20g at a speed of 500m/s. find the speed of the gun’s recoil and the energy of the explosion in the gun. 5. Two particles A and B of mass 3kg and 5kg respectively, collide. Just before impact A is moving with speed 8m/s and B is moving with speed 4m/s in the opposite direction. Immediately after impact A rebounds with speed 7m/s. find the speed of B immediately after impact. 6. A railway truck of mass 3000kg moving at 4m/s collides with a truck of mass 2000kg moving at 1m/s. if the trucks move on together after impact, find their common speed just after impact. If a constant retarding force of 400N acts on the trucks after impact, find the distance they travelled before coming to rest. Composition and resolution of forces and velocities. 1. Resultant force Forces and velocities can be added by using the triangular law or the parallelogram law. The triangular law: consider the forces given below. F R F RN P P The parallelogram law F R P Example1. find the resultant of two forces of magnitudes 4 N and 5 N which act at right angle to each other. 4N RN  5N We have from pythagoras’ theorem R  4  5  41 2 2 2  R  41  6.4 4 tan   0.8    tan 0.8  38.7 1 5 Example 2. two forces acting on a particle have magnitudes 2 N and 3N. If the angle between their directions is 600 , find the resultant force on the particle and its direction. Exercise 1. two forces of magnitudesFN and PN act on a block of wood such that the angle between them is . If RN is the resultant of the forces, solve for the unknown on each of the following Cases: (i) R  10 N , P  3N ,  20 ;0 (II) F  15 N , P  30 N ,  380 , (III) F;  35 N , P  23N , R  42 N (IV) F  76 N , P  60 N ,  120 , 0 (V) F  48 N , P  36 N ,  1020 ; (VI).R  18.3N , P  15.7 N , R  26.5 N Resolution of forces Any force can be resolved into two mutually perpendicular directions. Consider the forces P and Q in the x  y plane as given below: y Q P Q sin  P sin    Q cos  P cos  x Thus the sum of the forces along the x and y directions are: X  P cos   Q cos  Y  P sin   Q sin  The resultant of the P and Q is 2 and its inclination to R  X Y 2 Y the x-axis is given by tan   X Example1. Resolve each of the following forces along the x and y direction and fin the resultant and its inclination to the positive x-axis. 50 N 0 y 20 N 60 30 0 x Resolving along the x direction gives  3   50  1 X  20 cos 30  50 cos 60  20 0 0   2  2  10 3  25  5(2 3  5) Resolving along the x direction gives 1  3 Y  20 sin 30  50 sin 60  20   50 0 0   2  2   10  25 3  5(2  5 3 ) The resultant is given by R  X Y 2 2 R 5(2   2  3  5)  5(2  5 3 )  10 29 2 Its inclination to the positive x direction is given by tan    Y 5( 2  5 3) X 5(2 3  5) (2  5 3 ) tan    6.94073 (2 3  5)    tan 1 (6.94073)  81.80 This angle is in the second quadrant. Thus the inclination of the resultant to the positive x-axis is   180  81.8014  98.1986   98.2 0 EXERCISE 1. Find the magnitudes of the horizontal and vertical components of (a) a force of 25 N acting at to the 40 0 horizontal, (b) a force of 30 N acting at 76 0 to the horizontal, (c) a force of 18 N acting at 54 to the vertical, 0 EXERCISE 2. (a) A set of coplanar forces of magnitudes 10 N ,12 N ,18 N and 14 N act on a particle in directions N 600W , south-west, North and north-west. Find the magnitude and direction of their resultant. (b) A particle is being pulled along a smooth horizontal plane by three strings. The directions of the strings are N 60 W ,due north and N 30 E. If the 0 0 tensions in the string are of magnitude 14 N ,11N and 9N respectively, find the resultant force on the particle and give its direction. Exercise 3. (i) A block of mass 3kg is suspended by means of two light wires such that one wire makes an angle of 30 with the 0 upward vertical and the other makes an angle of 30 with the horizontal. 0 Given that the wires are on the same plane and that the block is in equilibrium find the tension in each String. (ii)A bead of mass 0.5kg is threaded to the mid point of a light inextensible string. Both ends of the string are fastened to the points A and B of two vertical pegs. Given the A and B are at a vertical height of 20cm above the bead and that the length of the string is 50cm find the tension on each string and its inclination to the horizontal when the bead is in equilibrium (iii) A wooden block of mass 10kg rests on a smooth plane which is inclined at an angle of 60 to the horizontal. Given 0 that the block is in equilibrium when a string which makes an angle of 15 with 0 the plane find the tension in the string. LAMI’S THEOREM This states that if three forces are in equilibrium they are concurrent and each force is proportional to the sin of the angle between the other two forces. P F  F P Q     sin  sin  sin  Q Example 1. A particle of mass 8kg is attached to an end of a light inelastic string of length 4m. The other end of the string is fastened to a fixed vertical wall. The particle is also held at a distance of 2.4m away from the wall by a horizontal force F. Find F and the magnitude of the tension in the string. Example 2. A smooth ring R of weight 7.5N is free to move on alight inextensible string which is fastened to two points A and B at the same horizontal level. The ring is kept in equilibrium by a horizontal force of magnitude F such that angle ABR is 30 0 and AR is vertical. Find the magnitude of the force F. Example 3. One end of a light inextensible string of length 72m is fixed to a vertical pole. A particle of eight 24N is attached to the other end of the string. The is pulled and held 24cm away from the pole by a horizontal force P. Find the magnitude of P and the tension in the string. Example 4. An object of weight 9N is suspended from a point in space by a light inextensible string. The object is in equilibrium by means of a horizontal force F when the string is inclined at an angle of 63 to the horizontal. Find the 0 magnitude of F and the tension in the string. Example 5. A particle of weight 65N is hanging in equilibrium supported by two light inextensible strings which are inclined at 72 and 58 to the 0 0 horizontal. Find the magnitudes of the tensions in the string. Example 6. A particle of weight 20N is held at rest on a smooth plane which is inclined at 30 to the horizontal by a 0 light string. Find the tension in the string and the reaction between the particle and the plane if (i) the string is parallel to the plane, (ii) the string is inclined at 60 to the plane. 0 VECTORS A vector is a quantity that has both magnitude and direction. A scalar quantity , on the other hand, has magnitude only but not direction. Example of vectors include the following: displacement; velocity; acceleration; weight; force. Examples of scalar quantities are: length; distance; mass; speed; energy; work and power. Representation of vectors The following are different   ways of B a representing vectors: ; b ; , ; A A A B b~ Vectors in two Dimensions We represent vectors in two dimensions as follows:  a  3 i  2 j, B  b i  b (i) a   , B    or (ii)  3   b 1 ~ ~ ~ 1 ~ 2 j ~ ~  2  b2  We can represent vectors on the Cartesian coordinate plane as follows: y r r  x i y j ~ ~ ~ y ~ j i ~ ~ x x We can add and subtract vectors in two dimensions. If a  3 i  2 j and b  5 i  4 j we have: ~ ~ ~ ~ ~ ~ (i) a  b  3 i  2 j  5 i  4 j  3  5  i  2  4  j  8 i  6 j  3 i  2 j   5 i  4  j  3  5  i  2  4  j   2 i  2 j ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ (ii) a  ~ b    ~   ~ ~ ~ ~ ~ ~ ~ ~ Exercise: Given the vectors a  4 i 6 j , ~ ~ ~ b  7 i  5 j c   10 i  11 j ~ ~, ~ , find the following: ~ ~ ~ (i) a b ; (ii) a b ; (iii) 3 a 2 c , (iv) 5 b 11 c , ~ ~ ~ ~ ~ ~ ~ ~ (v) 10 a 7 c ~ ~ We can also express velocity, force and acceleration in component form along the i and j directions. Example: If forces ~ ~ of magnitude 20 N and30 N act on a particle in directions making angles of 300 and1200 respectively with the positive x direction find their resultant in vector form and give its magnitude and y direction. 30 N j 20 N ~ 0 60 300 i x ~ We resolve the 20 N force as follows: F  20 cos 30 i  20 sin 30 j ~1 ~ ~  3 1   20  i  20  j   2  2 ~ ~  10 3 i  10 j..............(1) ~ ~ We resolve the 30 N force as follows: F  30 cos 60 i  30 sin 60 j ~ 2 ~ ~ 1  3  30  i  30 j ~  2~  2   15 i  15 3 j..............(2) ~ ~ The resultant force is given in vector form as R  F  F ~ ~1 ~ 2  10 3 i  10 j  15 i  15 3 j ~ ~ ~ ~      10 3  15 i  10  15 3 j ~ ~ R  52 3  3i  52  3 3  j ~ ~ ~ The magnitude of the resultant force is R  52 3  3  52  3 3  2 2 given by ~     25 12  12 3  9  25 4  12 3  27   5 12  12 3  9  4  12 3  27  5 12  9  4  27  5 52  20 13 N The inclination of the resultant to the positive x direction is Y    tan 1   X    tan  1  5 23 3   tan 1 23 3    86.31 0  5 2 3 3    2 3 3   Exercise: If forces of magnitude 10 N and 15 N act on a particle in directions making angles of and respectively 45 0 150 0 with the positive x direction find their resultant in vector form and give its magnitude and direction. Exercise1 four light inelastic strings are attached to a block of wood of weight 10N. All the strings act in a vertical plane through the centre of the block and make angles of 30 ,133 ,210 and300 0 0 0 0 with the horizontal on the same side of the block. Express the resultant of all the forces acting on the block in vector form and find its magnitude and inclination to the horizontal. VECTORS IN THREE DIMENSIONAL SPACE We can represent vectors in the three dimensional Cartesian coordinate system as shown Z below: P z k r ~ ~ Y x i O j ~ ~ X y From the figure above we have OP  r  x i  y j  z k ~ ~ ~ ~ OP is the position vector of the point P relative to the origin O. thus the following points 3,4,5, 2,6,1,and 1,9,7 have the position vectors given bya  3 i  4 j  5 k ~ ~ ~ ~ , b~  2 ~i  6 j  k~ and c~  ~i  9 ~i  7 ~i ~ We can also extend the concept of addition and subtraction to the case of three dimensional vectors as follows: Given the vectors a~  3 ~i  4 ~j  5 k~ , b  8 i  6 j  10 k then ~ ~ ~ ~ (i) a  b  3 i  4 j  5 k  8 i  6 j  10 k ~ ~ ~ ~ ~ ~ ~ ~  3  8  i  4  6  j  5  10  k ~ ~ ~  11 i  10 j  15 k ~ ~ (ii) ~ a  b  3 i  4 j  5 k   8 i  6 j  10 k  ~ ~ ~ ~ ~  ~ ~ ~   3  8  i  4  6  j  5  10  k ~ ~ ~  5 i 2 j 5 k ~ ~ ~ THE MODULUS OR MAGNITUDE OF A VECTOR Referring the above figure, we represent the modulus or magnitude of OP  r  x i  y j  z k as OP  r  x  y  z ~ ~ ~ ~ 2 2 2 Example 1. Given the vectors m  4 i  8 j  2 k , n  6 i  5 j  2 k find the ~ ~ ~ ~ ~ ~ ~ ~ magnitude of each of the following vectors: (i) m n,(ii) m n , (iii) 2~m 3 n~ , (iv) ~ ~ ~ ~ 4 n 3 m. ~ ~ THE SCALAR PRODUCT (DOT PRODUCT) If a~ and b are two vectors, their dot ~ product is given by a b  a b cos where  ~ ~ ~ ~ is the angle between a~ and b. ~ Example consider the vector forces given below and find their dot product. a 10 N ~ 30 0 a b  a b cos  (10)(15) cos300 ~ ~ ~ ~ b 15 N 3 ~  150  75 3 2 If a~  a1 ~i  a2 ~j  a3 k~ and b~  b1 i~  b2 ~j  b3 k~    a b   a1 i  a2 j  a3 k    b1 i  b2 j  b3 k   ~ ~  ~ ~ ~  ~ ~ ~  a1b1 i. i  a1b2 i. j  a1b3 i. k  a2b1 j. i  a2b2 j. j ~ ~ ~ ~ ~ ~ ~ ~ ~ ~  a2b3 j. k  a3b1 k. i  a3b2 k. j  a3b3 k. k ~ ~ ~ ~ ~ ~ ~ ~  a1b1  a2b2  a3b3 Since ~i. ~i  ~j. ~j  k~. k~  1 and i. j  j.i  i.k  k.i  j.k  k. j  0 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ We thus have a b  a b cos   a1b1  a2b2  a3b3 ~ ~ ~ ~ a b a1b1  a2b2  a3b3  cos   ~ ~  ab a12  a22  a32 b12  b22  b32 ~ ~  a1b1  a2b2  a3b3  1      cos  a 2  a 2  a2 b2  b2  b2   1 2 3 1 2 3  Example 1. find the angle between each of the following vectors: (i) a~  4 ~i  3 j  2 k~ , b~  5 ~i  j  2 k~ ~ ~ (ii) a~  i~  3 j  2 k~ , b~  4 ~i  j  6 k~ ~ ~ (iii) a~  7 ~i  3 j  k~ , b~  9 i~  3 j  3 k~ ~ ~ Exercise 1) Show that the vectors given by a  4 i  10 j  5 k , b  5 i  j  2 k are ~ ~ ~ ~ ~ ~ ~ perpendicular to each other. ~ 2) Find the value of s for which the vectors a~  ~i  ~j  5 k~ , b~  5 ~i  s ~j  k~ are perpendicular. 3) The forces F~ 1  4 ~i  3 j  2 k~ , F~ 2  5 ~i  ~j  2 k~ ~ and F  10 i  6 j  8 k act on a wooden ~ 3 ~ ~ ~ block. Find the fourth F which must be ~ 4 applied on maintain the block in equilibrium.