2. CP Vector Components

3. Scalars and Vectors A quantity is something that you measure. Scalar quantities have only size, or amounts. Ex: mass, temperature, volume, love, etc. Vector quantities have both a size and a direction (such as N, S, E or W) Ex: displacement, velocity, acceleration

4. Vectors are symbolized with arrows A

5. A vector has components. If the components are on the axes they are called rectangular components. The sum of a vector’s components equals the vector. X component of A or A x Y component of A or A y A

6. Usually, using the components is easier. A vector and its components are interchangeable. You can either use the vector or its components, depending on which is easier.

7. Trig functions are used to calculate vector components SOH CAH TOA

8.  A Consider this triangle… Hypotenuse Opposite Side (also A y ) Adjacent Side (also A x ) Finding Vector Components

9. Be careful, the x component is not always the cos function. The y component is not always the sin function.  A sin  A cos  A

10. Calculate the x and y components of the following: 25 o 13 m/s X Y X = 5.5 m/s Y = 12 m/s

11. Calculate the x and y components of the following: 18 o 20. m X Y X = 19 m Y = 6.2

12. Drawing the resultant vector from components (the parallelogram method) 1. Draw your vectors tail to tail 2. Sketch out a parallelogram that has sides equal to your original vectors. 3. Draw in the resultant from the tails to the opposite corner (going through the parallelogram diagonally)

13. Calculating the resultant vector To calculate the resultant vector, use the Pythagorean theorem. a 2 + b 2 = c 2 The Pythagorean theorem only works for right triangles. Therefore, rectangular (x & y) components must be used.

14. Describing the angle/direction: Using components to calculate a resulting vector

15. Draw and calculate the resultant vector: 8.0 N 3.0 N 8.0 N 3.0 N R 2 = (8.0) 2 + (3.0) 2 R 2 = 73 R = 8.5

16. Calculate the angle using inverse tangent Use inverse tangent to calculate the angle.  = tan -1 (opposite/adjacent)

17. Calculate the angle 8.0 N 3.0 N 8.0 N 3.0 N R 2 = (8.0) 2 + (3.0) 2 R 2 = 73 R = 8.5 N  Final answer: 8.5 N at 69 o _____

18. Describe the angle using north, south, east, and west. The direction of the vector used for the “opposite side” is named first. The direction of the vector used for the “adjacent side” is named last E North of East East of North North of West  N S W  N S E W  N S E W

19. Calculate the angle 8.0 N 3.0 N 8.0 N 3.0 N R 2 = (8.0) 2 + (3.0) 2 R 2 = 73 R = 8.5 N  Final answer: 8.5 N at 69 o E of S

20. Adding Vectors in Real Life… Step 1: Draw a Vector Diagram A=10.0 m 20. o B=15 m 30. o C=10. 0 m Find The Sum of A + B + C

21. Adding Vectors in Real Life… Step 2: Create data table holding x and y components of each vector and the total x and y components of the resultant vector. A=10.0 m 20. o B=15 30. o C=10.0 m

22. A=10.0 m 20. o B=15. m 30. o C=10.0 m 10.0cos20 10.0sin20 15sin30 15cos30 X X 9.397 m3.420 m 7.5 m- 13.0 m 0-10.0 m XY C B A

23. Step 3: Add the vectors along each axis to get the total resultant x and y components. 9.397 m3.420 m 7.5 m-13.0 m 0-10.0 m XY C B A Total -3.6 m0.92 m Remember: When adding you round to the least amount of decimal places (but don’t round until the end!)

24. 0.92 m 3.6 m Step 4: Draw a Vector Diagram showing only the vector axis sums from step 3. I dropped the negative sign because the arrow is pointing in the negative x direction

25. 0.92 m 3.6 m R 2 =(3.6) 2 + (0.92) 2 R = 3.7 m Step 5: Use the Pythagorean Theorem (a 2 + b 2 = c 2 ) to find the magnitude of the resultant vector. R = 4 m

26. 3.6 m R  Step 6a: Use a trig function (usually tan) to find the angle. 10 o 0.92 m

27. 3.6 R  R = 4 m @ 10 o North of West Step 6b: Specify both magnitude and direction of the vector. 0.92 m

28. Wow! That’s so much work!

29. 25.0 m 10.0 m 20. o 25cos20 o = 23.5 25sin20 o = 8.55 X Example: Add the vectors below. 23.5 m-8.55 m 0 m-10.0 m XY TOTAL B A 13.5 m-8.55 m

30. Example: Add the vectors below. 8.55 m 13.5 m R R 2 = 13.5 2 + 8.55 2 =16 m Tan  = 8.55/13.5  = 32. o  16 m, 32 o south of east.