Chap 1.pdf

VECTORS IN Chapter 1 TWO DIMENSIONS RESULTANT OF PERPENDICULAR VECTORS REVISION OF IMPORTANT CONCEPTS FROM GRADE 10 In Grade 10 you were acquainted with scalars and vectors. Let’s do...

VECTORS IN Chapter 1 TWO DIMENSIONS RESULTANT OF PERPENDICULAR VECTORS REVISION OF IMPORTANT CONCEPTS FROM GRADE 10 In Grade 10 you were acquainted with scalars and vectors. Let’s do a quick revision of the most important concepts regarding scalars and vectors. Scalar quantity Vector quantity Scalars are physical quantities, Vectors are physical quantities, pos- consisting of magnitude only. sessing magnitude and direction. Certain scalars are numbers only (numerical magni- A vector quantity is indicated by a number (numerical tudes), such as 1, 2, 3, etc. without units. In natural magnitude), a unit and a direction. For instance, a velo- sciences we specify scalars more completely with a city of 4 m.s-1 to the right, a force of 20 N Easterly and a number and the applicable unit such as a mass of 20 displacement of 10 m upwards kg, a velocity of 5 m.s-1 or a charge of 20 C. We can add and subtract scalar magnitudes, just as we When adding vectors, we have to take the direction of can do with ordinary numbers. For example, a distance each vector into account. When two people push a crate of 14 m + 13 m = 27 m and a velocity of 20 m.s-1 - 4 in the same direction, the effect is not the same as when m.s-1 = 16 m.s-1. they are pushing in opposite directions. VECTOR QUANTITIES SCALAR QUANTITIES Physical Quantity SI unit Symbol for Physical Quantity SI unit Symbol for quantities symbol of measurement unit quantities symbol of measurement unit displacement x meter m time t seconds s velocity v meter per second m.s-1 mass m kilogram kg acceleration a meter per second squared m.s-2 distance D meter m power F newton N speed v meter per second m.s-1 weight w newton N charge Q coulomb C Graphic presentation of vectors A vector quantity is indicated by a straight line with an arrow point. magnitude tail Parts of a vector head - The length of the line represents the magnitude of the vector magnitude. - The arrow head represents the direction of the vector magnitude. - The starting point of the vector is called the tail, and the end, indicated by the arrow head, is called the head. - The magnitude and direction of a vector can be represented more accurately when plotted to scale. Scale: 1 cm = 10 N N 50 N East The graphic representation of a vector is called a vector diagram. Symbolic representation of scalars and vectors Letters of the alphabet are used for both scalars and vectors. However, because vectors (which have magnitude and direction) differs from scalars (which only have magnitude), their notations differ. Scalars are indicated by ordinary letters, e.g. m for mass and D for distance. Vectors are indicated by bold letters, e.g. F for force, or in ordinary letters with an arrow directly above it, → e.g. F for force. The magnitude of a vector is written in ordinary letters as in the case of scalars. E.g. F for the magnitude of a force. MECHANICS 1 Direction of vectors left right In Grade 10 we represented vectors in one dimension (in a straight line). West East The utilization of positive and negative vectors is useful when working with North vectors in one dimension. Where one direction is indicated as positive, the opposite direction has to up be indicated as negative. We indicate opposite Horizontal direction (left and right, or East and West): Where right or East directions as "+" and "-". is regarded as “positive”, left or West must be regarded as “negative”. Vertical direction (up and down or North and South): Where up or North is regarded as “positive”, down or South must be regarded as “negative”. down In Grade 11 we study vectors in two dimensions. Therefore vectors can be denoted in various ways. There are three methods we can use to South denote direction. 0° (North) (1) Bearings D A A line of reference is chosen as north, with an angle of 0˚. All angles are then measured directly by clockwise angular rotation (0˚ - 360˚) from the 30° reference line. Figure 1 shows how this method can be utilised to indicate the following 320° C directions: 270° 120° A = direction 30° or bearing of 30° B = direction 120° or bearing of 120° B C = direction 270° or bearing of 270° Figure 1 D = direction 320° or bearing of 320° (2) Compass readings In this method we utilise the directions of wind, north, south,east and west. Figure 1 shows how this method can be imple- North C mented to indicate the following directions: A = 60° east of north, or N 60° O 30° B = 20° west of south, or S 20° W 60° A C = 30° west of north, or N 30° W (3) x- and y-axes West East When the x- and y-axes are used to indicate direction, the angle is measured in relation to the x-axis. It is also imperative to indicate whether the angle is above or underneath the x-axis, and whether the angle was measured in relation to the positive or negative x-axis. Figure 2 20° Figure 3 shows how this method can be implemented to indicate the following B directions: South A = 30° above the positive x-axis B = 45° underneath the positive x-axis y C = 37,5° underneath the negative x-axis A Addition of vectors Two or more vectors can be added together graphically or algebraically to 30° produce a single vector as answer, which has the same effect as the two or -x x more separate vector magnitudes. This combined effect of two or more vectors 37,5° 45° is called the resultant (of resultant vector R). C The resultant of a number of vectors is the single vector which has B the same effect as the original vectors combined. -y Figure 3 The resultant of two or more forces F1 and F2 can be indicated as a vector equation: FR = F1 + F2. Remember: Both the magnitude and the direction of each vector have to be taken into account. Two forces F1 of 100 N east and F2 of 70 N east are algebraically calculated as FR = F1 + F2 = 100 N + 70 N = 170 N east. The following figure indicates how these two force vectors can be added together graphically by sketching the force vector F2 with its tail connected to the head of force vector F1. This graphical method of vector addition is called the tail-to-head method. F1 = 100 N F2 = 70 N east as positive FR = 170 N east 2 TOPIC 1 When forces are exerted in opposite directions, they are still added together algebraically. With the equation F1 100 N east but F2 altered to 70 N west, the easterly direction is now indicated as positive; the westerly direction is then negative. F2 is in the opposite direction as F1 and thus F2 is then called the negative vector so that: FR = F1 + F2 = (+100 N) + (-70 N) = 30 N east. The following diagram indicates how these two force vectors can be added together graphically by drawing the force vector F2 with its tail connected to the head of force vector F1, but in the opposite direction. F1 = 100 N east as positive FR = 30 N east F2 = 70 N We always indicate the resultant vector by the number representing its magnitude, the correct unit and the correct direction. Exercise 1 REVISION OF IMPORTANT CONCEPTS OF GRADE 10 1. Explain the concepts and provide three examples of each: scalar quantities and vector quantities Scalar: __________________________________________________________________________________ ______________________________________________________________________________________________ Vector : __________________________________________________________________________________ ______________________________________________________________________________________________ ______________________________________________________________________________________________ Take 0˚ (North) as reference line. Use lines 2. 3. Indicate the wind directions on the following measuring 4 cm with arrow heads to indicate axes system. Use lines measuring 4 cm with the following directions on the same axes sys- arrow heads to indicate the following directions tem: on the same axes system: A = direction 45° ; B = direction 110° A = 60° W of N ; B = South-East C = direction 180° ; D = direction 330° C = 35° S of W ; D = 55° N of O 4. Describe each of the directions A, B and C in Question 2, as well as A,C and D in Question 3 in terms of an angle measured in relation to the positive or negative x-axis. For Question 2: For Question 3: A = ______________________________________ A = ______________________________________ B = ______________________________________ C = ______________________________________ C = ______________________________________ D = ______________________________________ MECHANICS 3 5. Consider the sketch: 5.1 Indicate the directions of A, B, C and D clockwise, measured from the 0˚ (North) reference line. 0° (north) A = _________________________________ A B = _________________________________ D 30° C = _________________________________ 65° D = _________________________________ 270° (west) 90° (east) 5.2 Indicate the directions of A, B, C and D in terms of a sharp angle and the basic wind direction. A = _________________________________ 37° 20° B B = _________________________________ C 180° (south) C = _________________________________ D = _________________________________ 6. What is meant by the resultant of vectors? _________________________________________________________________________________________ ________________________________________________________________________________________________ 7. A force F1 of 5 N and a force F2 of 3 N are simultaneously exerted on an object. The direction of the forces can be altered at will. 7.1 When will these forces have their greatest resultant? ___________________________________________ ______________________________________________________________________________________________ 7.2 Confirm your answer in 7.1 by determining the resultant of these forces, utilizing an accurate tail-to-head vector diagram. Use a scale of 10 mm : 1 N. 7.3 Test your answer in Question 7.2 by using a vector equation. _____________________________________________________________________________________ 7.4 When will these forces have their least resultant? __________________________________________ ________________________________________________________________________________________ 7.5 Confirm your answers in Question 7.4 by determining the resultant of these forces, utilizing an accurate tail- to-head vector diagram. Use a scale of 10 mm : 1 N 7.6 Test your answers in Question 7.5 by using a vector equation. _____________________________________________________________________________________ 4 TOPIC 1 8. Determine the resultant force graphically or by calcu- 60 N 30 N 40 N 70 N lation,of all the forces during a tug-of-war competition as 50 N illustrated in the sketch. Use a scale of 10 mm:10 N. 9. The following forces are exerted on an object: FA = 4,5 N downwards; FB = 3,2 N downwards en FC = 11,5 N upwards. Determine the resultant of the three forces by utilizing a head-to-tail vector diagram. Test your answer by utilizing a vector equation. 10.Two forces, F1 and F2 are exerted on a crate as illustrated. F1 is 15,5 N to F1 = 15,5 N F2 the right. Use a vector diagram as well as a vector equation to calculate F2 if the resultant of the two forces is 7,3 N is. MECHANICS 5 THE RESULTANT OF TWO PERPENDICULAR VECTORS y Cartesian co-ordinate system Vectors can be plotted on a Cartesian plane. The Cartesian plane is a set of axes, crossing each other perpendicularly at the origin (0;0). The horizontal axis (x- axis) is positive to the right and negative to the left of the origin; the vertical axis (y-axis) is positive upwards of the origin and negative below the origin. -x x 0 Representation of vectors on the Cartesian plane origin Rx is a vector in the horizontal direction. Ry is a vector in the vertical direction. The two vectors are perpendicular (90˚) in relation to one another. Two methods can be utilised to represent these two perpendicular vectors on a Cartesian plane, namely by plotting them tail-to-head or by plotting them -y tail-to tail. A Cartesian plane. Method 1: Tail-to-head Method 2: Tail-to-tail Starting at the origin, plot vector Rx in the x-direction. Starting at the origin, plot vector Rx in the x-direction. Plot vector Ry in the y-direction from the head of Rx, so Plot vector Ry in the y-direction from the origin, so that that the two vectors are plotted tail-to-head. the two vectors are plotted tail-to-tail. y Ry Ry y -x x -x x 0 Rx 0 Rx -y -y Determine the resultants R or Rx and Ry The resultant vector R of two or more vectors is the single vector which has the same effect as the original vectors together. Two methods can be used to determine the resultant vector R of two vectors Rx and Ry which act upon each other perpendicularly, namely the tail-to-head method and the tail-to-tail method. I. Graphical determination of the resultant vector 1. Tail-to-head method 2. Tail-to-tail method Obtain a suitable scale, e.g. Scale: 10 mm : 10 N. Obtain a suitable scale, e.g. Scale: 10 mm : 10 N. Plot the first vector, Rx according to scale. Plot the first vector, Rx according to scale. Plot the second vector, Ry according to scale, Plot the second vector, Ry according to scale, start at attaching its tail to the head of Rx the same origin as Rx (i.e. start at the tail of Rx). Connect the tail of Rx with the head of Ry to produce Complete a rectangle by plotting two lines of equal the resultant vector R. length parallel to Rx en Ry Measure the length of R with a ruler and the Plot the resultant vector R as the diagonal of the direction (θ) of R with a protractor. rectangle, starting at the tails of Rx and Ry. Use the chosen scale to determine the true mag- Measure the length of R with a ruler and the direction nitude of the resultant. (θ) of R with a protractor. Describe the direction (θ) according to the methods Use the chosen scale to determine the true magnitude you have learnt to indicate direction. of the resultant. Describe the direction (θ) of R according to the The tail-to-head method used to add two methods you have learnt to indicate direction. vectors together, is often referred to as the triangular method because the third side of y the triangle is formed by the resultant The tail-to-tail method used to add two vectors together, is referred to as the parallelogram R R method, as it forms a Ry Ry parallelogram y θ θ -x x -x x 0 Rx 0 Rx -y -y 6 TOPIC 1 II. Determination of the resultant vector by calculation 1. Tail-to-head method 2. Tail-to-tail method Plot a sketch-diagram (i.e. a vector dia- Plot a sketch-diagram (i.e. a vector dia- gram which is not according to scale) with gram which is not according to scale) with vectors Rx and Ry tail-to-head. vectors Rx and Ry tail-to-tail Plot the resultant vector R. Plot the resultant vector R. Indicate the angle θ between Rx and R. Indicate the angle θ between Rx and R. y We use R when refer- ring to a resultant vector in general. FR is used R more specifically to re- y Ry R fer to a remaining force. Ry -x 0 θ x θ Rx -x 0 x Rx -y -y Magnitude of the resultant vector R Utilise Pythagoras’ Theorem to obtain the magnitude of the resultant vector R: R2 = Rx2 + R2y Direction of the resultant vector R Utilise trigonometry to obtain the direction of the resultant vector R: Ry Describe the direction (θ) of R tan θ = __ according to the methods you Rx learned to indicate direction In Grade 11 we use examples which include force vectors and displacement vectors to graphically calculate the magnitude and direction of the resultant. Let us look at two computed examples: Example 1 Force vectors Two forces, F1 of 80 N east and F2 of 40 N north are simultaneously exerted on a wooden barrel as illustrated in the sketch. F2 = 40 N north (1) Plot a sketch-diagram (i.e. a vector diagram which is not according to scale) of the force vectors on a Cartesian plane. (2) Using a scale of 10 mm : 10 N and the tail-to-tail method. Calcu- late the magnitude and direction of the resultant of these two forces graphically, F1 = 80 N east Answer (1) y (North) (2) Scale: 10 mm :10 N (I.e.: 1 mm:1 N) F2 = 40 N north y 4 F2 = 40 N north -x x FR 3 100 1 90 10 0 F1 = 80 N east 80 7 0 12 0 60 130 50 -y 2 14 0 3 0 4 15 θ 0 0 2 160 0 cm 1 0 10 170 180 -x x 0 F1 = 80 N east Magnitude: FR = [89 (mm) × 1 N] = 89 N 0 cm 1 2 3 4 5 6 7 8 Direction: θ = 27° above the positive x-axis -y The magnitude and direction of the resultant force: FR = 89 N, 27° above the positive x-axis (or, 27° north of east or E 27° N) (or, direction / bearing 63°) MECHANICS 7  Example 2 Displacement vectors Susan walks 5 km in the direction 0˚ (in the positive y-direction), where after she chan- ges course and then walks 8 km in the direction 90˚ (in the positive x-direction) (1) Plot a sketch-diagram (i.e. a vector diagram not according to scale), to indicate the two displacements Susan underwent, as well as her resultant displacement, using the tail-to-head method. label all information completely in the sketch. (2) Calculate the magnitude and direction of Susan’s resultant displacement. Answer (1) y (2) Magnitude of R R2 = R2y + Rx2 Direction of R Ry tan θ = __ Rx = 8 km = (5 km)2 + (8 km)2 Rx = 25 km2 + 64 km2 8m Ry = 5 km = 89 km2 = ___ 5m R = 9,4 km R = 1,6 θ Susan’s resultant displacement, θ = tan-1(1,6) R = 9,4 km direction 58° -x x θ = 58° (or, 58° east of north or N 58° E) -y (or, 32° above the positive x-axis Exercise 2 RESULTANT OF TWO PERPENDICULAR VECTORS 1. Explain the following concepts: Explain the following concepts : _____________________________________________________________ ______________________________________________________________________________________________ _________________________________________________________________________________________ ______________________________________________________________________________________________ Perpendicular vectors : ____________________________________________________________________ ______________________________________________________________________________________________ ______________________________________________________________________________________________ Tail-to-head method : _______________________________________________________________________ ______________________________________________________________________________________________ ______________________________________________________________________________________________ ______________________________________________________________________________________________ Tail-to-tail method : ________________________________________________________________________ ______________________________________________________________________________________________ ______________________________________________________________________________________________ ______________________________________________________________________________________________ ______________________________________________________________________________________________ 8 TOPIC 1 2. Two forces of 120 N each are exerted on a crate simultaneously as shown in 120 N the figure. 2.1 Plot a sketch diagram of the force vectors on a Cartesian plane. 120 N 2.2 Using a suitable scale of 10 mm : 20 N and the tail-to-tail method determine the magnitude and direction of the resutants of these two forces graphically. 3. Force vectors P and Q are plotted according to scale on the Cartesian plane shown below.. Calculate: 3.1 the magnitude of vector P, in force units. y-axis __________________________________________ 4 Vector P __________________________________________ 3 __________________________________________ 2 __________________________________________ 1 -4 -3 -2 -1 1 2 3 4 x-axis __________________________________________ 0 -1 3.2 the direction of vector Q, measured clockwise from the Vector Q positive y-axis -2 -3 __________________________________________ -4 __________________________________________ __________________________________________ __________________________________________ __________________________________________ MECHANICS 9 4. Two Newton spring balances are used to exert forces on a wooden block as shown in the figure. 4. 1 Take the readings of the individual forces exerted on the spring balances N and plot them on a Cartesian plane. 1 2 3 4 5 6 7 8 9 10 10 N 6 7 2 5 4 9 8 3 1 4.2 Plot a vector sketch diagram (not according to scale) 4.3 Calculate the magnitude and direction of the two forces and their resultant. of the resultant of the two forces. ____________________________________ ____________________________________ ____________________________________ ____________________________________ ____________________________________ ____________________________________ ________________________________________ ________________________________________ ________________________________________ ________________________________________ ________________________________________ ________________________________________________________________________________________ 5. A boy on his surfboard moves 150 m due west, after which he changes direction, moving 70 m due north. Plot the two displacements on a Cartesian plane and then calculate the boy’s resultant displacement relative to the horizontal plane. ________________________________________ ____________________________________ ________________________________________ ____________________________________ ________________________________________ ____________________________________ ________________________________________ ____________________________________ ________________________________________ ____________________________________ ________________________________________________________________________________________ 10 TOPIC 1 6. While being rowed across a river 800 m wide, a boat is swept 500 m downs team before reaching the opposite bank. Determine the resultant displacement of the boat by means of an accurate scale diagram. Utilise the head- to-tail method. 7. A horizontal and a vertical force is exerted on an object. The horizontal force is established as being 8 N. Graphi- cally determine the magnitude of the vertical force, as well as the angle between the resultant and the horizontal force, if the magnitude of the resultant is 10 N. 8. A horizontal and a vertical force are exerted on an object. The vertical force is established as being 5 N. Calculate the magnitude of the horizontal force, as well as the angle between the resultant and the horizontal force, if the magnitude of the resultant is 9 N. MECHANICS 11 THE RESULTANT OF MORE THAN TWO I. ADDITION OF COLLINEAR VECTORS Two or more vectors operating in one and the same dimension (i.e. A B E appearing on the same straight line), are called collinear vectors. Collinear vectors can be aligned either in the same or in opposite C D directions. In the accompanying figure the vectors are all collinear vectors, except vector E. A ,B, C and D are collin- We are now going to learn how to represent collinear horizontal vectors ear horizontal vectors. and collinear vertical vectors on a Cartesian plane and how to determine A and B are collinear vectors in the the resultant of these vectors. same direction. C and D are collinear vectors in the oppsite direction. Representation of collinear vectors on the Cartesian plane When two or more vectors simultaneously work in the horizontal x-direction, we can plot these vectors on the Cartesian plane with the head-to-tail method; thus they represent the collinear horizontal vectors. In Figure 1 below, Fx1 and Fx2 are two force vectors (collinear forces), working simultaneously in the horizontal x-direction. y Figure 1 Figure 2 y Figure 3 y Fx1 Fx2 -x 0 x Fy2 Rx = Fx1 + Fx2 Fy2 Net or resultant horizontal Ry = Fy1 + Fy2 force vector Rx. Fy1 Fy1 -x x -x x 0 Fx1 Fx2 0 Collinear forces represented Net or resultant vertical on a Cartesian plane. force vector Ry. In the same way, when two or more vectors are simultaneously working in the vertical y-direction, we can plot these vectors head-to-tail on the Cartesian plane and thus they represent the collinear vertical vectors. In Figure 1 above, Fy1 and Fy2 are two force vectors (collinear forces), working simultaneously in the vertical y-direction. Add the collinears together The advantage of collinear vectors is that they can be added together algebraically. Now we join together the collinear horizontal vectors by adding up all vectors in the horizontal direction (i.e. parallel in the x-direction) to determine the net or resultant horizontal vector Rx , (also known as the net x-component), as indicated in Figure 2 above. The net horizontal force vector or net x-component of Fx1 and Fx2 then equals: Rx = Fx1 + Fx2 Likewise we add the vertical collinear vectors together by adding all the vectors in the vertical direction (ie perpendicular to the x-direction) to determine the net or resultant vertical vector Ry (also called the net y-component), as in Figure 3 above. The net vertical force vector of net y-component of the two forces Fy1 and Fy2 is: Ry = Fy1 + Fy2 Sketch Rx and Ry on the Cartesian plane After determining the net horizontal vector (net x-component) Rx and the net vertical vector (net y-component) Ry , we can plot these two vectors on a new Cartesian plane, as illustrated in the figure below. There are two methods we can use to plot the two vectors Rx and Ry on the Cartesian plane. y Method 1: Tail-to-tail method Method 2: Tail-to-head method Starting at the origin, plot vector Rx in Starting at the origin, plot vector Rx in the x-direction. Plot vector Ry also from the x-direction. Plot vector Ry in the y- the origin in the y-direction in order to direction from the head of Rx in order to plot the vectors tail-to-tail. get the two vectors tail-to-head. Ry Ry y -x x -x x 0 Rx 0 Rx -y -y 12 TOPIC 1 Determine the resultant R of Rx and Ry The resultant vector R of two or more vectors is the single vector which has the same effect as the original vec- tors together. There are two methods to determine the resultant vector R of two perpendicular vectors Rx and Ry : the tail-to-head method and the tail-to-tail method. I. Determining the resultant vector graphically 1. Tail-to-head method 2. Tail-to-tail method Obtain a suitable scale, e.g. Scale: 10 mm : 10 N. Obtain a suitable scale, e.g. Scale: 10 mm : 10 N. Plot the first vector, Rx according to scale. Plot the first vector, Rx according to scale. Plot the second vector, Ry according to scale, Plot the second vector, Ry according to scale, start at attaching its tail to the head of Rx the same origin as Rx (i.e. start at the tail of Rx). Connect the tail of Rx with the head of Ry to pro- Complete a rectangle by plotting two lines of equal duce the resultant vector R. length parallel to Rx en Ry Measure the length of R with a ruler and the Plot the resultant vector R as the diagonal of the direction (θ) of R with a protractor. rectangle, starting at the tails of Rx and Ry. Use the chosen scale to determine the true mag- Measure the length of R with a ruler and the direction nitude of the resultant. (θ) of R with a protractor. Describe the direction (θ) according to the methods Use the chosen scale to determine the true magnitude you have learnt to indicate direction. of the resultant. Describe the direction (θ) of R according to the methods you have learnt to indicate direction. The tail-to-head method used to add two vectors together, is often referred to as the triangular method because the third side of y the triangle is formed by the resultant The tail-to-tail method used to add two vectors together, is referred to as the parallelogram R R method, as it forms a Ry Ry y parallelogram θ θ -x x -x x 0 Rx 0 Rx -y -y II. Determining the resultant vector algebraically 1. Tail-to-head method 2. Tail-to-tail method Plot a sketch-diagram (i.e. a vector dia- Plot a sketch-diagram (i.e. a vector dia- gram which is not according to scale) with gram which is not according to scale) with vectors Rx and Ry tail-to-head. vectors Rx and Ry tail-to-tail Plot the resultant vector R. Plot the resultant vector R. Indicate the angle θ between Rx and R. Indicate the angle θ between Rx and R. y R R y Ry Ry -x 0 θ x -x 0 θ x Rx Rx -y -y Magnitude of the resultant vector R Utilise Pythagoras’ Theorem to obtain the magnitude of resultant R: R2 = Rx2 + R2y Direction of the resultant vector R Utilise trigonometry to obtain the direction of resultant vector R: Ry Describe the direction (θ) of R tan θ = __ according to the methods you Rx learned to indicate direction. We now look at the following examples which include force vectors and displacement vectors to determine the magnitude and direction of the resultant of collinear horizontal vectors and collinear vertical vectors. MECHANICS 13  Example 3 Force vectors The figure shows six toy tractors, all six simultaneously exert- F4 = 50 N ing a force on a crate. Three horizontal forces: F1 = 35 N east, F2 = 25 N east en F3 = 30 N west, and three vertical forces: F4 F5 = 30 N = 50 N north, F5 = 30 N north and F6 = 40 N south. (1) Plot a sketch diagram of the forces on a Cartesian plane F1 = 35 N (2) Plot a diagram of the net vertical force (net y-component) Fy and the net horizontal force (net x-component) Fx. Plot the resultant FR of Fy and Fx on the diagram. (Use the tail- to-tail method). (3) Calculate the magnitude and direction of the resultant FR F3 = 30 N of the forces. F2 = 25 N Answer F6 = 40 N y (1) (2) Fx = F1 + F2 + F3 Fy = F4 + F5 + F6 F5 = 30 N = 35 N + 25 N + (-30 N) = 50 N + 30 N + (-40 N) = 30 N east = 40 N north y F4 = 50 N FR 0 -x x Fy = 40 N F3 = 30 N F1 = 35 N F2 = 25 N θ F6 = 40 N -x x 0 Fx = 30 N -y -y (3) Magnitude of FR Direction of FR FR2 2 = Fy + Fx 2 Fy tan θ = __ = (40 N)2 + (30 N)2 Fx 40 N = 1 600 N2 + 900 N2 = ____ (or direction (bearing of) 36,9°) 30 N = 2 500 N2 (or E 53,1° N or N 36,9° E) = 1,33 FR = 50 N (or, 53,1° above the positive x-axis) θ = tan-1(1,33) θ = 53,1° The magnitude and direction of the resultant force: FR = 50 N, 53,1° north from east Example 4 Displacement vectors A boy walks 10 m due north, then 3 m due east, stands still for a moment and then walks another 2 m due east, then 4 m due south and a further 13 m due west. (1) Plot a sketch diagram of the boy’s displacement on a Cartesian plane. (2) Calculate net displacement (net component) of the boy’s displacement in the direc- tion north-south (vertical). (2) Calculate net displacement (net component) of the boy’s displacement in the direc- tion east-west (horizontal). (3) Plot an accurate vector diagram according to scale (10 mm = 1 m) of the boy’s net displacement to the north and his net displacement to the south. Determine the magnitude and direction of his displacement in respect of the point of origin by ac- curate measurement. (Use the tail-to-head method). 14 TOPIC 1  Answer (1) y (North) (2) Net displacement vertical Fy (or vertical) = R1 + R5 = 10 m + (-4 N) = 6 m north R1 = 10 m north (3) Net displacement horizontal Fx (or horizontal) = R2 + R3 + R4 0 R3 = 2 m east = 3 m + 2 m +(-13 m) -x x R4 = 13 m west = -8 m R2 = 3 m east I.e.: 8 m west R5 = 4 m south -y (4) Scale: 10 mm = 1 m 6 Magnitude: R = [100 (mm) × 1 m] = 100 m Direction: θ = 36° north of west 5 Fy (or vertical) = 6 m north The magnitude and direction of the re- 4 R sultant displacement from the starting point: R = 89 N, 36° north of west 3 001 09 011 0 21 07 08 1 03 06 05 2 04 1 y (North) 4 2 0 10 θ 5 10 3 Fx (or horizontal) 081 071 06 1 (or, W 36° N or N 54° W) 0 01 0 = 8 m west -x x (or direction (bearing of) 306°) 0 0 0 1 2 3 4 5 6 7 8 (or, 36° above the positive x-axis) -y Exercise 3 ADDITION OF COLLINEAR VECTORS 1. Explain the concept of collinear vectors. _________________________________________________________________________________________ ______________________________________________________________________________________________ 3. The figure shows six forces exercised simultaneously on an object. Three horizontal forces: F1 = 40 N east, F2 = 50 N east and F3 = 30 N east, and three vertical forces: F4 = 40 N north, F5 = 35 5 N south and F6 = 45 N west. 3.1 Plot a sketch diagram of the forces on a Cartesian plane. F4 = 40 N F1 = 40 N F3 = 30 N F2 = 50 N F5 = 35 N F6 = 45 N MECHANICS 15  2.2 Plot a sketch diagram (not to scale) of the vertical force (net y-component) Fy and the net horizontal force (net x-component) Fx. Draw the resultant FR of Fy and Fx on the diagram. (Use the tail-to-tail method.) 2.3 Calculate the magnitude and direction of the resultant FR of the forces. Magnitude: ______________________________________ Direction: __________________________ ______________________________________ __________________________ ______________________________________ __________________________ ______________________________________ __________________________ _____________________________________________________________________________________ 3. Force vectors F1, F2, F3, F4, F5 and F6 are shown according scale on the Cartesian plane below. 50 3.1 Calculate the net force (net component) of 40 all the forces in the horizontal x-direction. F4 ___________________________ 30 20 ___________________________ 10 F1 ___________________________ -40 -30 -20 -10 F2 x-as F3 0 10 20 30 40 50 60 70 80 90 100 ___________________________ -10 3.2 Calculate the net force (net component) of -20 all the forces in the vertical y-direction. -30 ___________________________ F5 -40 ___________________________ -50 F6 ___________________________ -60 -70 ___________________________ 3.3 Plot a sketch diagram of the net horizontal force (net x-component) Fx and the net vertical force (net y- component) Fy. Draw the resultant FR of Fx and Fy on the diagram. (Use the tail-to-head method.) 16 TOPIC 1  3.4 Calculate the magnitude and direction of the resultant FR of the forces. Magnitude: ______________________________________ Direction: __________________________ ______________________________________ __________________________ ______________________________________ __________________________ ______________________________________ __________________________ _____________________________________________________________________________________ 4. A yacht sailing on the following route from the port. N 5 km north, 2 km north, 3 km east, W O 2 km east and then 4 km south 4.1 Plot a sketch diagram upon a Cartesian plane S of the displacement of the yacht 4.2 Calculate net displacement (net component) of 4.3 Calculate net displacement (net component) of the yacht in the north-south direction (vertical) the yacht in the east-west direction (horizontal) ______________________________________ ______________________________________ ______________________________________ ______________________________________ ______________________________________ ______________________________________ ______________________________________ ______________________________________ 4.4 Plot a accurate vector diagram to scale (10 mm:1 m) of the yacht's net displacement north and its net dis- placement south. By accurate measurement determine the magnitude and direction of the yacht's displace- ment with respect to the port (starting point). (Use the tail-to-head method.) MECHANICS 17 II. ADDITION OF VECTORS WHICH ARE NOT COLLINEAR OR PERPENDICULAR Occasionally several vectors can exist, e.g. forces acting on an object from different directions or multiple displacements can occur in different directions. The resultant vector of any number of vectors on a 2 dimensional plane (2D- plane) is then calculated by the polygon method of tail-to-head addition. The first arrow is plotted from a suitable starting point to represent one of the vectors according to a suitable scale. A second arrow is plotted from its end to represent the second vector. This process is repeated until each of the vectors is represented by an arrow. The answer will not be influenced by the sequence (consecutive order) of the illustrated vectors. The resultant is indicated by the arrow plotted from the starting point (tail) of the first arrow to the end (head) of the last arrow. The magnitude and direction of the resultant of the vectors are determined by accurate measurement. This method applies to all kinds of vectors (e.g. force, displacement, velocity, etc.). It can also be used to determine the resultant of only two vectors. Example 5 Force vectors Four forces, consisting of F1 = 4 N, F2 = 5 N, F3 = 6 N en F4 = 5 N are F1 = 4 N, 0° exerted at the same point on an object, in directions 0˚, 110˚, 220˚ and F4 = 5 N, 290° 290˚ respectively. Determine the vale and direction of the resultant of the forces with the aid of an accurate scale drawing. F2 = 5 N, 110° Answer Utilise the polygon method of tail-to-head addition. Start with a small axes system and plot according to scale an arrow of 4 F3 = 6 N, 220° cm in the direction 0˚ to represent F1 in magnitude and direction. Plot a new small axes system at the point (head) of F1. From this axes system, and according to scale, plot an arrow of 5 cm in the direction 110˚ to represent F2 and both its magnitude and direction. Plot a new small axes system at the point (head) of F2. From this axes system, and according to scale, plot an arrow of 6 cm in the direction 220˚ to represent F3 and both its magnitude and direction. Plot a new small axes system at the point (head) of F3. From this axes system, and according to scale, plot an arrow of 5 cm in the direction 290˚ to represent F4 and both its magnitude and direction. Now connect the beginning (tail) of F1 with the end (head) of F4 , with the arrow pointing in the direction of F4. This arrow (vector) which completes the polygon, is the resultant force FR of the four forces exerted on the object. (FR = F1 + F2 + F3 + F4). Determine the magnitude and direction of FR by accurate measurement. Scale: 1 cm = 1 N 0° Magnitude of FR 0° 110° F2 = 5 N FR = [3,9 cm) × 1 N] = 3,9 N F1 = 4 N FR 220° 4 5 3 2 1 0 FR Direction of FR θ F3 = 6 N 15 0 160 170 180 10 0 20 1 40 3 0 40 60 130 50 0 12 F4 = 5 N 10 0 100 1 80 7 290° FR 90 100 θ 80 110 The resultant force, FR = 3,9 N, direction 260° 70 2 0 1 60 3 0 1 50 (or 10° under the negative x-axis) 40 01 0 θ = (180° + 80°) 6 0 15 4 180 170 1 30 (or 10° South of West, of W 10° S) 0 0 10 2 = 260° (or 80° West of South, or S 80° W) 18 TOPIC 1 Example 6 Displacement vectors A boat starts navigating the following route from the harbour: 62 km in the direction 90˚ ; 15 km in the direction 180˚ ; 77 km in the direction 315˚ and then 30 km in the direction 240˚. Using a scale of 10 mm = 10 km, determine: (1) the boat’s resultant displacement; (2) the direction in which the boat has to navigate in a straight line to return to the harbour. Answer (1) Utilise the polygon method of tail-to-head addition Start with a small axes system and, according to scale, and draw an arrow of 6,2 cm (62 mm) in the direction 90˚ to indicate the magnitude and direction of the displacement. Plot a new axes system at the end (head) of the arrow. From the new axes system, draw an arrow of 1,5 cm (15 mm) in the direction 180˚ according to scale, to represent the displacement of 15 km. Plot a new axes system at the end (head) of the arrow. From the new axis system, draw an arrow of 7,7 cm (77 mm) in the direction 180˚ according to scale, to represent the displacement of 7 km. Plot a new axes system at the end (head) of the arrow. From the new axes system, draw an arrow of 3 cm (30 mm) in the direction 240˚ according to scale, to represent the displacement of 30 km. Now connect the start point (tail) of the first arrow with the end (head) of the last arrow, in the direction of the last arrow. This arrow (vector), which completes the polygon, is the resultant vector R of the four displacements the boat underwent. Determine the magnitude and direction of R by accurate measurement. Scale: 10 mm = 10 km Magnitude of R 240° 0° 30 km 0 1 R α 2 77 km 0° 3 FR = [30 cm) × 10 N] R = 30 N 4 180° 62 km Direction of R 0° θ 15 km 315° 11 001 09 10 08 02 07 06 05 1 04 03 1 2 0 1 04 5 3 10 The resultant displacement, R = 30 km, direction 320° FR 081 071 06 0 01 0 (or 50° above the negative x-axis) (or 50° North of West, or W 40° N) θ = (270° + 50°) θ (or 40° West of North, or N 40° W) = 320° (2) Direction to navigate back to the harbour = α = direction 140° Direction (α) α = (90° + 50°) = 140° (or 50° under the negative x-axis) (or 50° South of East, of E 50° S) α 90 (or 40° East of South, or S 40° E) 100 1 80 7 0 10 60 130 R 12 0 50 Note: If the end of the last displacement coincides with the beginning 4 14 0 3 0 2 of the starting point of the first displacement, the figure is closed. In 0 15 0 10 0 0 160 such a case it is a zero resultant or zero displacement. The boat is 170 180 then back at he starting point (harbour). MECHANICS 19  Closed polygon of forces If the result of the vector addition by way of the tail-to-head method is a closed polygon, then the resultant force FR = 0 N and the forces are in equilibrium (balanced). Such an object will stay at rest or move at a constant velocity. F1 Figure 2 F2 FR = F1 + F2 + F3 + F4 = 0 Figure 1 F1 F4 F3 F2 F3 F4 The four forces in equilibrium is represented in sequence by the sides of a polygon. They form a polygon of forces. Figure 1 above represents the force vectors of four known forces, exerted on an object simultaneously. The polygon in Figure 2 is the vector diagram obtained by placing the vectors tail-to-head, in sequence (in magnitude and direction). The vector diagram is a closed polygon which means that the four forces do not have a resultant force. FR = F1 + F2 + F3 + F4 = 0 If three or more forces do not constitute a closed triangle or polygon, it means that there is a resultant force. The vector that has to be plotted to complete the figure, will represent the resultant force if its direction is not measured in the same sequence. When a triangle or polygon of forces constitute a closed figure with its sides in sequence, the resultant force is zero, in other words the forces exerted on the same point, are in equilibrium (are balanced). Exercise 4 ADDITION OF VECTORS WHICH ARE NOT COLLINEAR OR PERPENDICULAR 1. Define resultant of two or more vectors in words. _________________________________________________________________________________________ ______________________________________________________________________________________________ 2. Name the method used to graphically determine the resultant of any number of forces (two, three, four or more) acting simultaneously in different directions on an object. ________________________________________________________________________________________ 3. Four forces of 80 N, 100 N, 110 N and 160 N act in at the same point (object) into directions of respectively 90 °, 45 °, 300 ° and 250 °. Use an accurate scale drawing where 10 mm = 20 N to determine the resultant (magnitude and direction) of these forces. 20 TOPIC 1 4. Forces of 4 N, 3 N, 9 N and 5 N are exerted simultaneously on a object as shown in the 4N figure. Use an accurate scale drawing to determine the resultant force (magnitude and direction) of all the forces acting on the object. 5N 3N 45° 9N 8,66 N 5. Four forces act in at the same object like show in the figure. 10 N 5.1 Use an accurate scale drawing to determine the resultant (magnitude and 30° direction) all the forces acting on the object. 45° 30° 5N 8,66 N MECHANICS 21  5.2 What does a vector of this nature reveal? _____________________________________________________________________________________ _____________________________________________________________________________________ _____________________________________________________________________________________ 6. A fisherman in a boat sails at first 4 km south from the port, later he sails 5 km east and then 3 km in the direction 45 ° N of E. Use an accurate scale diagram (where 1 km is represented by 10 mm) in order to determine the following: 6.1 How far is the angler from the harbor after all the movements? 6.2 In which direction must he sail in order to get to the port again via a straight line? 7. A long distance runner ran a route that consists of four legs. These are: 10 m east, 16 m in the direction 30 °, 20 m east of 30 ° south and 20 m in the direction 300 ° 7.1 What is the total distance ran by the athlete? ________________________________ 7.2 Use a scale (10 mm = 2 m) and draw an accurate scale diagram to the represent route the athlete traveled. Determine from your diagram the athlete's resultant displacement. 22 TOPIC 1 RELATIONSHIP BETWEEN THREE NON-LINEAR FORCES IN EQUILIBRIUM FORCES IN EQUILIBRIUM When an object is stationary (or is moving at a constant velocity) it is possible (1) that no forces are exerted on it, or (2) that the forces exerted on it are balanced , in other words the F1 forces are cancelled to deliver a zero resultant. We say these cancelled forces are in equilibrium. When the resultant of two or more forces exerted F2 on a point (object) are ZERO, the forces are in equilibrium (balanced). FR = 0 In the figure an stubborn donkey are held by Jack and Ronald so F1 that no movement occurs. Two forces, F1 and F2 , are exerted by the men. The resultant of these forces, FR , is balanced exactly by Resultant, FR Equilibrant, FE a force F3 , which is exerted by the donkey. Force F3 which equals FR , but is exerted in the opposite direction of FR , is called the equilibrant (that which keeps the balance) FE. FE = -FR F2 The equilibrant (that which keeps the balance) of two or more forces is the single force which keeps the other forces in equilibrium (balance). The equilibrant has the same magnitude as the resultant of the forces, but operates in the opposite direction. When three forces are exerted at a point (object) and it stays stationary, or it moves at a constant velocity, the forces are in balance and the resultant force equals zero. A head-to-tail diagram of the three forces shows that they form a triangle with no resultant. It is called the Triangle rule of three forces in Equilibrium and is defined thus: When the forces exerted at a point (object) are in equilibrium, their in magnitude and direction can be represented sequentially by the sides of a triangle. [“Sequentially” is intended to indicate they are plotted “consecutively”] As the three forces F1, F2 and F3 , which are exerted on a point, are in equilibrium, a head-to-tail diagram of these forces will produce a closed triangle. F1 + F2 + F3 = 0 F2 F3 F2 F1 F1 F3 RESULTANT OF THREE NON-LINEAR FORCE VECTORS The figure on the right indicates the setup of an apparatus which is used in an experiment to determine the resultant of three non-linear F3 θ2 θ1 F2 forces. Two pulleys are attached to a force board. A sheet of white paper is attached to the board. Three pieces of string are tied to a ring. Weights are tied to the other end of each string and they are draped F1 over the pulleys in such a way that they can move freely; the ring are then adjusted vertically so that they are in the approximate centre of the sheet of paper. The three forces exerted on the ring, are balancing one another (the system is in equilibrium). The centre of the ring is plotted on the paper, as are the directions in which forces F1, F2 and F3 are exerted. The magnitude of the three forces is determined by calculating the weight of each one in N (newton) with w (Fg) = mg = (mass in kg)(9,8 m.s-2). MECHANICS 23  The mass of each weight is noted on the paper – i.e. the magnitude of each force. Figure 1 The paper is removed from the board. The forces are marked as in figure 1 and the angles θ1 and θ2 are measured in relation to the vertical line. F3 F2 The tail-to-head method is implemented to calculate the F3 θ1 θ2 resultant of the three forces F1, F2 and F3. θ2 We plot the vertical force first, followed tail-to-head by the F1 other two forces. Ensure that the angle from the vertical line to each force is correct. Discussion F2 The system of strings and weights is at rest, and the three θ1 F1 forces acting on the ring are in balance. The resultant should be zero. The tail-to-head vector addition of the three forces Tail-to-head vector dia- constitute a closed triangle (the last vector’s head is at the gram of the three forces tail of the first vector). The resultant is zero. This confirms that the forces are balanced (forces are in equilibrium). It confirms the Triangle rule of three forces in Equilibrium. Plot the three forc- es on the paper Note! The resultant of TWO vectors can also be determined R tail-to-tail with a parallelogram of forces (parallelogram method), which reads as follows: resultant of F2 and F3 When TWO forces are exerted on a point, these two S forces can be represented in magnitude and direction