1. NEWTON’S LAWSVECTORS PHY1012F VECTORS Gregor Leigh gregor.leigh@uct.ac.za

2. NEWTON’S LAWSVECTORS PHY1012F 2 VECTORS Learning outcomes: At the end of this chapter you should be able to… Resolve vectors into components and reassemble components into a single vector with magnitude and direction. Make use of unit vectors for specifying direction. Manipulate vectors (add, subtract, multiply by a scalar) both graphically (geometrically) and algebraically.

3. NEWTON’S LAWSVECTORS PHY1012F 3 VECTORS Using only positive and negative signs to denote the direction of vector quantities is possible only when working in a single dimension (i.e. rectilinearly). In order to deal with direction when describing motion in 2-d (and later, 3-d) we manipulate vectors using either graphical (geometrical) techniques, or the algebraic addition of vector components.

4. NEWTON’S LAWSVECTORS PHY1012F 4 SCALARS and VECTORS Scalar – A physical quantity with magnitude (size) but no associated direction. E.g. temperature, energy, mass. Vector – A physical quantity which has both magnitude AND direction. E.g. displacement, velocity, force.

5. NEWTON’S LAWSVECTORS PHY1012F 5 VECTOR REPRESENTATION and NOTATION Graphically, a vector is represented by a ray. The length of the ray represents the magnitude, while the arrow indicates the direction. Algebraically, we shall distinguish a vector from a scalar by using an arrow over the letter,. Note: r is a scalar quantity representing the magnitude of vector, and can never be negative. The important information is in the direction and length of the ray – we can shift it around if we do not change these.

6. NEWTON’S LAWSVECTORS PHY1012F 6 GRAPHICAL VECTOR ADDITION A helicopter flies 15 km on a bearing of 10°, then 20 km on a bearing of 250°. Determine its net displacement. N 10°  15 km 20 km  = 74° 60°

7. NEWTON’S LAWSVECTORS PHY1012F 7 MULTIPLYING A VECTOR BY A SCALAR Multiplying a vector by a positive scalar gives another vector with a different magnitude but the same direction: Notes: B = cA. (c is the factor by which the magnitude of is changed.) lies in the same direction as. (Distributive law). If c is zero, the product is the directionless zero vector, or null vector.

8. NEWTON’S LAWSVECTORS PHY1012F 8 VECTOR COMPONENTS Manipulating vectors geometrically is tedious. Using a (rectangular) coordinate system, we can use components to manipulate vectors algebraically. We shall use Cartesian coordinates, a right-handed system of axes: y x z (The (entire) system can be rotated – any which way – to suit the situation.)

9. NEWTON’S LAWSVECTORS PHY1012F 9 VECTOR COMPONENTS Adding two vectors (graphically joining them head-to- tail) produces a resultant (drawn from the tail of the first to the head of the last)…

10. NEWTON’S LAWSVECTORS PHY1012F 10 VECTOR COMPONENTS Adding two vectors (graphically joining them head-to- tail) produces a resultant (drawn from the tail of the first to the head of the last)… “Running the movie backwards” resolves a single vector into two (or more!) components.

11. NEWTON’S LAWSVECTORS PHY1012F 11 VECTOR COMPONENTS Adding two vectors (graphically joining them head-to- tail) produces a resultant (drawn from the tail of the first to the head of the last)… “Running the movie backwards” resolves a single vector into two (or more!) components.

12. NEWTON’S LAWSVECTORS PHY1012F 12 VECTOR COMPONENTS Adding two vectors (graphically joining them head-to- tail) produces a resultant (drawn from the tail of the first to the head of the last)… “Running the movie backwards” resolves a single vector into two (or more!) components. Even if the number of components is restricted, there is still an infinite number of pairs into which a particular vector may be decomposed. Unless… ??!

13. NEWTON’S LAWSVECTORS PHY1012F 13 VECTOR COMPONENTS …by introducing axes, we specify the directions of the components. y x is now constrained to resolve into and, at right angles to each other. Note that, provided that we adhere to the right-handed Cartesian convention, the axes may be orientated in any way which suits a given situation.

14. NEWTON’S LAWSVECTORS PHY1012F 14 VECTOR COMPONENTS Resolution can also be seen as a projection of onto each of the axes to produce vector components and. y x A x, the scalar component of (or, as before, simply its component) along the x- axis … has the same magnitude as. remains unchanged by a translation of the axes (but is changed by a rotation). is positive if it points right; negative if it points left.

15. NEWTON’S LAWSVECTORS PHY1012F 15 VECTOR COMPONENTS The components of are… y (m) 2 4 0 6 8 24608 x (m) A x = +6 m A y = +3 m

16. NEWTON’S LAWSVECTORS PHY1012F 16 0 VECTOR COMPONENTS The components of are… y (m) 2 4 6 8 -6-4-2-8 x (m) A x = +6 m A y = +3 m

17. NEWTON’S LAWSVECTORS PHY1012F 17 VECTOR COMPONENTS The components of are… y (m) -6 -4 -2 -6-4-2-8 x (m) A x = +6 m A y = +3 m -8

18. NEWTON’S LAWSVECTORS PHY1012F 18 VECTOR COMPONENTS The components of are… y (m) -2 2 4 -4-22 x (m) A x = –2 m A y = +4 m

19. NEWTON’S LAWSVECTORS PHY1012F 19 VECTOR COMPONENTS The components of are… y (m) -2 2 4 -8-44 x (m) A x = –6 m A y = –5 m

20. NEWTON’S LAWSVECTORSPHY1012F 20 VECTOR COMPONENTS The components of are… y (m) 2 4 -8 -4 4 x (m) A x = –6 m A y = –5 m -4 –8 m –3 m

21. NEWTON’S LAWSVECTORS PHY1012F PHY1010W 21 VECTOR COMPONENTS The components of are… y (m) x (m) A x = +A cos  m A y = – A sin  m 

22. NEWTON’S LAWSVECTORS PHY1012F PHY1010W 22 VECTOR COMPONENTS The components of are… y (m) x (m) A x = –A sin  m A y = – A cos  m  Note that we can (re)combine components into a single vector, i.e. (re)write it in polar notation, by calculating its magnitude and direction using Pythagoras and trigonometry: (On this slide!)

23. NEWTON’S LAWSVECTORS PHY1012F 23 UNIT VECTORS Components are most useful when used with unit vector notation. A unit vector is a vector with a magnitude of exactly 1 pointing in a particular direction: y x 1 1 A unit vector is pure direction – it has no units! Vector can now be resolved and written as:

24. NEWTON’S LAWSVECTORS PHY1012F 24 UNIT VECTORS y (m/s) x (m/s) 60° v x = –(12 m/s) cos60°  v x = –6.00 m/s v x = –v cos60° v = 12 m/s v y = +v sin60° v y = +(12 m/s) sin60°  v y = +10.4 m/s Hence: Given a 12 m/s velocity vector which makes an angle of 60° with the negative x-axis, write the vector in terms of components and unit vectors.

25. NEWTON’S LAWSVECTORS PHY1012F 25 ALGEBRAIC ADDITION OF VECTORS Suppose Thus D x = A x + B x + C x and D y = A y + B y + C y In other words, we can add vectors by adding their components, axis by axis, to determine a single resultant component in each direction. These resultants can then be combined, or simply presented in unit vector notation. and

26. NEWTON’S LAWSVECTORS PHY1012F 26 y x + ALGEBRAIC ADDITION OF VECTORS The process of vector addition by the addition of components can visualised as follows: AxAx BxBx ByBy AyAy D y = A y + B y D x = A x + B x =

27. NEWTON’S LAWSVECTORS PHY1012F 27 ALGEBRAIC ADDITION OF VECTORS While it is often quite acceptable to present as y x D y = A y + B y D x = A x + B x its polar form is easily reconstituted from D x and D y using and 

28. NEWTON’S LAWSVECTORS PHY1012F 28 GRAPHICAL VECTOR ADDITION A helicopter flies 15 km on a bearing of 10°, then 20 km on a bearing of 250°. Determine its net displacement. N 10°  15 km 20 km  = 74° 60°

29. NEWTON’S LAWSVECTORS PHY1012F 29 ALGEBRAIC ADDITION OF VECTORS A helicopter flies 15 km on a bearing of 10°, then 20 km on a bearing of 250°. Determine its net displacement. N 10°  15 km 20° x- comp’nt (km) y- comp’nt (km) y x 20 km +15 sin10°+15 cos10° –20 cos20°–20 sin20° –16.2+7.93 +2.60+14.77 –18.79–6.84

30. NEWTON’S LAWSVECTORS PHY1012F 30 N y x ALGEBRAIC ADDITION OF VECTORS A spelunker is surveying a cave. He follows a passage 100 m straight east, then 50 m in a direction 30° west of north, then 30° 45° 50 m 100 m 150 m 150 m at 45° west of south. After a fourth unmeasured displacement he finds himself back where he started. Determine the magnitude and direction of his fourth displacement.

31. NEWTON’S LAWSVECTORS PHY1012F 31 R x = 100 – 25 –106 + D x = 0 ALGEBRAIC ADDITION OF VECTORS N 30° 45° y x 50 m 100 m 150 m VectorMagntd (m) Angle  x- comp’nt (m) y- comp’nt (m) 1000°0° 50 120° 150 225° R x = 0 R y = 0 R y = 0 +43.3 –106 + D y = 0  D x = 31  D y = 62.7 1000 –25 43.3 –106 3162.7??69.9 63.7  ??

32. NEWTON’S LAWSVECTORS PHY1012F 32 VECTORS Learning outcomes: At the end of this chapter you should be able to… Resolve vectors into components and reassemble components into a single vector with magnitude and direction. Make use of unit vectors for specifying direction. Manipulate vectors (add, subtract, multiply by a scalar) both graphically (geometrically) and algebraically.