Vectors day 2

1. Vectors day 2

2. y j i x k z Unit vector notation (i,j,k) Consider 3D axes (x, y, z) Define unit vectors, i, j, k Examples of Use: 40 m, E = 40 i 40 m, W = -40 i 30 m, N = 30 j 30 m, S = -30 j 20 m, out = 20 k 20 m, in = -20 k

3. R +40 m f -30 m Example 4:A woman walks 30 m, W; then 40 m, N. Write her displacementini,jnotation and inR,qnotation. In i,j notation, we have: R = Rxi + Ry j Rx = - 30 m Ry = + 40 m R = -30 i + 40 j Displacement is 30 m west and 40 m north of the starting position.

4. R +40 m f -30 m Example 4 (Cont.):Next we find her displacementinR,qnotation. q= 1800 – 59.10 q = 126.9o R = 50 m (R,q) = (50 m, 126.9o)

5. 46 km f=? B 35 km R = ? A q = 1800 + 52.70 q = 232.70 Example 6:Town A is 35 km south and 46 km west of Town B. Find length and direction of highway between towns. R = -46 i – 35 j R = 57.8 km f = 52.70 S. of W.

6. F = 240 N Fy F 280 Or in i,j notation: Fy F = -(212 N)i+ (113 N)j Fx Example 7. Find the components of the 240-N force exerted by the boy on the girl if his arm makes an angle of 280 with the ground. Fx= -|(240 N) cos 280|= -212 N Fy= +|(240 N) sin 280|= +113 N

7. F = 300 N 32o 32o Fx 320 Fy Fy F Or in i,j notation: F = -(254 N)i- (159 N)j Example 8. Find the components of a 300-N force acting along the handle of a lawn-mower. The angle with the ground is 320. Fx= -|(300 N) cos 320|= -254 N Fy= -|(300 N) sin 320|= -159 N

8. 1. Start at origin. Draw each vector to scale with tip of 1st to tail of 2nd, tip of 2nd to tail 3rd, and so on for others. Component Method 2. Draw resultant from origin to tip of last vector, noting the quadrant of the resultant. 3. Write each vector in i,j notation. 4. Add vectors algebraically to get resultant in i,j notation. Then convert to (R,q).

9. N B 3 km, W 4 km, N C E A D 2 km, E 2 km, S Example 9.A boat moves 2.0 km east then 4.0 km north, then 3.0 km west, and finally 2.0 km south. Find resultant displacement. 1. Start at origin. Draw each vector to scale with tip of 1st to tail of 2nd, tip of 2nd to tail 3rd, and so on for others. 2. Draw resultant from origin to tip of last vector, noting the quadrant of the resultant. Note: The scale is approximate, but it is still clear that the resultant is in the fourth quadrant.

10. N B 3 km, W 4 km, N C E A D 2 km, S 2 km, E 5. Convert to R,q notation See next page. Example 9 (Cont.)Find resultant displacement. 3.Write each vector ini,jnotation: A = +2 i B = + 4 j C = -3 i D = - 2 j 4.Add vectors A,B,C,D algebraically to get resultant ini,jnotation. -1 i + 2 j R = 1 km, west and 2 km north of origin.

11. Resultant Sum is: R = -1 i + 2 j N B 3 km, W 4 km, N C E D A 2 km, S 2 km, E Ry= +2 km R f Rx = -1 km Example 9 (Cont.)Find resultant displacement. Now, We Find R,  R = 2.24 km  = 63.40 N or W

12. N For convenience, we follow the practice of assuming three (3) significant figures for all data in problems. D 3 km 2 km C B 4 km E A 2 km Reminder of Significant Units: In the previous example, we assume that the distances are 2.00 km, 4.00 km, and 3.00 km. Thus, the answer must be reported as: R = 2.24 km, 63.40 N of W

13. 30 lb R q 40 lb Ry f q Rx 40 lb R 30 lb Rx Ry Significant Digits for Angles Since a tenthof a degree can often be significant, sometimes a fourth digit is needed. Rule:Write angles to the nearest tenth of a degree. See the two examples below: q = 36.9o; 323.1o

14. A = 5 m, 00 C = 0.5 m R B = 2.1 m, 200 B q C = 0.5 m, 900 200 A = 5 m B = 2.1 m Example 10: Find R,q for the three vector displacements below: 1. First draw vectors A, B, and C to approximate scale and indicate angles. (Rough drawing) 2. Draw resultant from origin to tip of last vector; noting the quadrant of the resultant. (R,q) 3. Write each vector in i,j notation. (Continued ...)

15. For i,j notation find x,y compo-nents of each vector A, B, C. C = 0.5 m R B q 200 A = 5 m B = 2.1 m Example 10: Find R,q for the three vector displacements below: (A table may help.)

16. A = 5.00 i + 0 j B = 1.97 i + 0.718 j C = 0 i + 0.50 j 6.97 i + 1.22 j Example 10 (Cont.): Find i,j for three vectors: A = 5 m,00; B = 2.1 m, 200; C = 0.5 m, 900. 4. Add vectors to get resultant R in i,j notation. R =

17. R = 6.97 i + 1.22 j Diagram for finding R,q: R q Ry 1.22 m Rx= 6.97 m Example 10 (Cont.): Find i,j for three vectors: A = 5 m,00; B = 2.1 m, 200; C = 0.5 m, 900. 5. Determine R,q from x,y: R = 7.08 m q = 9.930 N. of E.

18. q Example 11:A bike travels 20 m, E then 40 m at 60o N of W, and finally 30 m at 210o. What is the resultant displacement graphically? C = 30 m Graphically, we use ruler and protractor to draw components, then measure the Resultant R,q B = 40 m 30o R 60o f A = 20 m, E R = (32.6 m, 143.0o) Let 1 cm = 10 m

19. Cy 30o R Ry 60o f 0 q Rx Ax Cx Bx A Graphical Understanding of the Components and of the Resultant is given below: Note: Rx = Ax + Bx + Cx By B Ry = Ay + By + Cy C A

20. Cy By 30o B C R Ry A 60 f q Rx Ax Cx Bx Example 11 (Cont.)Using the Component Method to solve for the Resultant. Write each vector in i,j notation. Ax = 20 m, Ay = 0 A = 20 i Bx = -40 cos 60o = -20 m By= 40 sin 60o = +34.6 m B = -20 i + 34.6 j Cx = -30 cos 30o = -26 m C = -26 i - 15 j Cy = -30 sin 60o = -15 m

21. Cy By 30o B C R Ry A 60 f q Rx Ax R= (-26)2 + (19.6)2 = 32.6 m Cx Bx R 19.6 -26 +19.6 tan f = f -26 Example 11 (Cont.)The Component Method Add algebraically: A = 20 i B = -20 i + 34.6 j C = -26 i - 15 j R= -26 i + 19.6 j q = 143o

22. Cy By 30o B C R Ry A 60 f q Rx Ax Cx Bx R +19.6 f -26 Example 11 (Cont.)Find the Resultant. R = -26 i + 19.6 j The Resultant Displacement of the bike is best given by its polar coordinates R and q. R = 32.6 m; q = 1430

23. Cx A = 5 m, 900 B Cy B = 12 m, 00 350 A C = 20 m, -350 C q R A = 0 i + 5.00 j B = 12 i + 0 j C = 16.4 i – 11.5 j 28.4 i - 6.47 j Example 12.Find A + B + C for Vectors Shown below. Ax = 0; Ay = +5 m Bx = +12 m; By = 0 Cx = (20 m) cos 350 Cy = -(20 m) sin -350 R =

24. B 350 A C q q R R Example 12 (Continued).Find A + B + C Rx = 28.4 m Ry = -6.47 m R = 29.1 m q = 12.80 S. of E.

25. First Consider A + B Graphically: B B R B A A Vector Difference For vectors, signs are indicators of direction. Thus, when a vector is subtracted, the sign (direction) must be changed before adding. R = A + B

26. Now A – B: First change sign (direction) of B, then add the negative vector. B B -B A R’ -B A A Vector Difference For vectors, signs are indicators of direction. Thus, when a vector is subtracted, the sign (direction) must be changed before adding.

27. Subtraction results in a significant difference both in the magnitude and the direction of the resultant vector. |(A – B)| = |A| - |B| Comparison of addition and subtraction of B B B A R R’ -B B A A Addition and Subtraction R = A + B R’ = A - B

28. A – B; B - A +A -A +B -B A 2.43 N B 7.74 N Example 13.Given A = 2.4 km, N and B = 7.8 km, N: find A – B and B – A. A - B B - A R R (2.43 N – 7.74 S) (7.74 N – 2.43 S) 5.31 km, S 5.31 km, N

29. Components of R: R Ry Rx = R cosq q Ry = R sin q Rx Summary for Vectors • A scalar quantity is completely specified by its magnitude only. (40 m, 10 gal) • A vector quantity is completely specified by its magnitude and direction. (40 m, 300)

30. Resultant of Vectors: R Ry q Rx Summary Continued: • Finding the resultant of two perpendicular vectors is like converting from polar (R, q) to the rectangular (Rx, Ry) coordinates.

31. Component Method for Vectors • Start at origin and draw each vector in succession forming a labeled polygon. • Draw resultant from origin to tip of last vector, noting the quadrant of resultant. • Write each vector in i,j notation (Rx,Ry). • Add vectors algebraically to get resultant in i,j notation. Then convert to (R,q).

32. Now A – B: First change sign (direction) of B, then add the negative vector. B B -B A R’ -B A A Vector Difference For vectors, signs are indicators of direction. Thus, when a vector is subtracted, the sign (direction) must be changed before adding.

33. Conclusion of Chapter 3B - Vectors