PHYS 20 LESSONS Unit 2: 2-D Kinematics Projectiles Lesson 1: 2-D Vectors Adding vectors (Tail-to-tip)

Reading Segment #1: 2-D Vectors To prepare for this section, please read: Unit 2: p.2

A. 2-D VECTORS Recall, when we dealt with 1-D vectors, there were only two possible directions. We made one direction positive and the other direction negative. Now, we will deal with 2-D vectors, where there are many more than two directions. Expressing its direction is more complicated.

A1. Direction of 2-D Vectors - there are two methods to express the direction of 2-D vectors that we use in Physics 20: 1. NEWS (North, East, West, South) - acute angles measured with respect to the nearest axis 2. RCS (rectangular coordinate system) - angles measured in standard position (from the positive x-axis)

1. Angles in NEWSN V  = 50  WE S Method:  The angle is always positive and less than 90  (acute)  The angle is described relative to a nearby axis i.e.  = 50  N of E (North of the East Axis)

e.g. V N 60  WE S How would express this angle in NEWS?

e.g. V N 60  WE S  = 60  of In the NEWS method: - you always show the acute angle on the diagram - then, you will place two letters after the angle (these are shown as spaces)

e.g. V N 60  WE S  = 60  of W The second letter is the axis that the angle "touches". In this case, the angle is touching the West axis.

e.g. V N 60  WE S  = 60  N of W The first letter is the direction the angle moves from the reference axis. In this case, the angle is moving towards the North.

e.g. V N 60  WE S  = 60  N of W We interpret this as: "60 degrees North from the West axis"

Ex. 1Express the following angles in NEWS. a)Nb) N W EW E 40  V 1 70  S SV 2

a)N WE V 1 70  S  1 = 70  of

a)N WE V 1 70  S  1 = 70  of S The second letter is the axis the angle touches. In this case, it touches the South axis.

a)N WE V 1 70  S  1 = 70  W of S The first letter is the direction the angle moves from the reference axis. In this case, it moves towards the West.

a)N WE V 1 70  S  1 = 70  W of S We interpret this as "70 degrees West from the South axis"

b)N WE 40  V 2 S  2 = 40  of

b)N WE 40  V 2 S  2 = 40  of E The second letter is the axis the angle touches. In this case, it touches the East axis.

b)N WE 40  V 2 S  2 = 40  S of E The first letter is the direction the angle moves from the reference axis. In this case, it moves towards the South.

b)N WE 40  V 2 S  2 = 40  S of E We interpret this as "40 degrees South from the East axis"

2. Angles in RCSy  -xx V-y Method:  The angle is measured in standard position - starts from the positive x-axis  Counterclockwise angles are positive Clockwise angles are negative

e.g. V y 60  -xx -y How would express this angle in RCS?

e.g. V y 60  start -xx -y RCS angles always start at the positive x-axis.

e.g. V y  60  -xx -y If the angle is counterclockwise, then it is positive. So,  = 180  - 60  = 120 

e.g. V y 60  -xx  -y But if the angle is clockwise, then it is negative. So,  = -90    = -240 

Ex. 2Express the following angles in RCS. a)yb) y -x x-x x 40  V 1 70  -y -yV 2

a)y Start -xx V 1 70  -y RCS angles start from the positive x-axis.

a)y -xx V 1 70   1 -y If the angle is clockwise, then it is negative. So,  1 = -90   = -160 

a)y  1 -xx V 1 70  -y If the angle is counterclockwise, then it is positive. So,  1 = 90  + 90  + 20  = 200 

b)y start -xx 40  V 2 -y RCS angles always start at the positive x-axis.

b)y -xx 40  V 2 -y Clockwise angles are negative. So,  2 = -40 

b)y  2 -xx 40  V 2 -y Counterclockwise angles are positive. So,  2 = -90     = -320 

Practice Problems Try these problems in the Physics 20 Workbook: Unit 1 p. 4 #1

Reading Segment #2: Adding 2-D Vectors (Tail-to-Tip) To prepare for this section, please read: Unit 2: p.3

B. ADDING 2-D VECTORS A key skill in Physics 20 and 30 is to add 2-D vectors. There are two methods that we use: 1. Tail-to-Tip - this is especially good when the vectors are perpendicular (at 90  ) 2. Components - a more tedious, labour-intensive method, but it works for all cases (including 3-D)

B1. Adding 2-D Vectors (Tail -to-Tip) Method:  Place the tail of the second vector on the tip of the first i.e. "place one right after the other"  The resultant (or sum) vector R is drawn from the origin to the tip of the second vector i.e. Resultant is the "start to finish" vector  Solve the resulting triangle If the vectors are at right angles, then you can use Soh Cah Toa and the Pythagorean formula

Note: Notice that the resultant vector is a "start-to-finish" vector. This is the same description for the overall displacement vector ( d ). Thus, displacement is a good example of adding 2-D vectors.

Ex. 3From a hunting lodge, a hiker walks the following path:  1.8 km South  then, 1.10 km East Find the overall displacement of the hiker. Include both magnitude and direction in your answer.

Strategy: Overall displacement is a start-to-finish vector, just like the resultant vector. So, we will add the vectors tail-to-tip to get out answer.

1.8 km 1.1 km Place the tail of the second vector onto the tip of the first vector. That is, place one vector right after the other.

 1.8 km R 1.1 km The resultant (displacement) is the "start-to-finish" vector. Place the angle  at the base (origin) of the resultant vector.

 1.8 km R 90  1.1 km Since it is a right triangle, we can use Soh Cah Toa and the Pythagorean formula to find R and .

Pythag:  c 2 = a 2 + b km R c a 1.1 km b Remember, c is always the hypotenuse (the longest side).

 c 2 = a 2 + b km R c a R 2 = (1.8 km) 2 + (1.1 km) km R 2 = 4.45 km 2 b R = 4.45 km 2 = 2.1 km

Soh Cah Toa:  1.8 km R hyp adj 1.1 km opp Remember: - the hypotenuse is the longest side - the opposite side is the side furthest from  (i.e. the one not touching the angle  ) - the adjacent side is the side right beside the angle 

Toa:  tan  = opp1.8 km R hyp adj adj 1.1 km opp Since we know the opposite and the adjacent sides, we will use tangent.

Toa:  tan  = opp1.8 km R hyp adj adj tan  = km 1.8 opp  = tan -1 (0.6111) = 31 

N WE 31  R = 2.1 km S The answer in NEWS: d = R = 2.1 km at 31  E of S (or 2.1 km at 59  S of E)

y  -xx 31  R = 2.1 km The answer in RCS: -y d = R = 2.1 km at 301  (or 2.1 km at -59  )

Animation: 2-D Addition (Tail-to-Tip) The second animation deals with forces, but it shows vector addition very well.

Ex. 4A boat takes the following course:  3.60 km West  then, 5.20 km North Find the overall displacement of the boat. Include both magnitude and direction in your answer.

5.20 km 3.60 km Place the tail of the second vector onto the tip of the first vector. That is, place one vector right after the other.

5.20 kmR  3.60 km The resultant displacement vector is the "start-to-finish" vector. Be certain to show the angle  at the base (start) of the resultant vector.

Pythag: c 2 = a 2 + b kmR R 2 = (3.60 km) 2 + (5.20 km) 2  3.60 km R = 40 km 2 = 6.32 km

Soh Cah Toa: tan  = opp 5.20 kmR adj  tan  = km 3.60  = tan -1 (1.4444) = 55.3 

N 6.32 km 55.3  WE S The answer in NEWS: d = R = 6.32 km at 55.3  N of W (or 6.32 km at 34.7  W of N)

y 6.32 km  55.3  -xx -y The answer in RCS: d = R = 6.32 km at 125  (or 6.32 km at -235  )

Ex. 5A person walks the following path:  3.0 km at 20  S of E  then, 5.0 km at 10  W of S Sketch the resultant displacement vector. No calculations required.

20  3.0 km Sketch the first vector: 3.0 km at 20  South from the East axis

20  3.0 km Next, place new axes at the tip of the vector.

20  3.0 km 10  5.0 km Add the second vector onto the tip of the first vector. 5.0 km at 10  West of the South axis

20  3.0 km R 10  5.0 km The resultant is the "start to finish" vector.

Practice Problems Try these problems in the Physics 20 Workbook: Unit 2 p. 4 #2 - 7