2. Chapter 1 – Math Review

3. 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. EN 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. 2 km, E A 4 km, N B 3 km, W C 2 km, S D

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

5. Example 9 (Cont.) Find resultant displacement. EN 2 km, E A 4 km, N B 3 km, W C 2 km, S D Resultant Sum is: R = -1 i + 2 j R y = +2 km R x = -1 km R  Now, We Find R,  R = 2.24 km  = 63.4 0 N or W

6. Reminder of Significant Units: EN 2 km A 4 km B 3 km C 2 km D For convenience, we follow the practice of assuming three (3) significant figures for all data in problems. 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, 63.4 0 N of W

7. Significant Digits for Angles 40 lb 30 lbR   RyRy RxRx 40 lb 30 lb R  RyRy RxRx  = 36.9 o ; 323.1 o Since a tenth of 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:

8. Example 10: Find R,  for the three vector displacements below: A = 5 m B = 2.1 m 20 0 B C = 0.5 m R A = 5 m, 0 0 B = 2.1 m, 20 0 C = 0.5 m, 90 0 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,  ) 3. Write each vector in i,j notation. (Continued...)

9. Example 10: Find R,  for the three vector displacements below: (A table may help.) Vector Vector  X-component (i) X-component (i) Y-component (j) Y-component (j) A=5 m A=5 m 00 00 00 00 + 5 m + 5 m 0 B=2.1m B=2.1m 20 0 +(2.1 m) cos 20 0 +(2.1 m) sin 20 0 +(2.1 m) sin 20 0 C=.5 m C=.5 m 90 0 0 + 0.5 m + 0.5 m R x = A x +B x +C x R x = A x +B x +C x R y = A y +B y +C y R y = A y +B y +C y A = 5 m B = 2.1 m 20 0 B C = 0.5 m R  For i,j notation find x,y compo- nents of each vector A, B, C.

10. Example 10 (Cont.): Find i,j for three vectors: A = 5 m,0 0 ; B = 2.1 m, 20 0 ; C = 0.5 m, 90 0. X-component (i) X-component (i) Y-component (j) Y-component (j) A x = + 5.00 m A x = + 5.00 m A y = 0 A y = 0 B x = +1.97 m B x = +1.97 m B y = +0.718 m B y = +0.718 m C x = 0 C x = 0 C y = + 0.50 m C y = + 0.50 m A = 5.00 i + 0 j A = 5.00 i + 0 j B = 1.97 i + 0.718 j B = 1.97 i + 0.718 j C = 0 i + 0.50 j C = 0 i + 0.50 j 4. Add vectors to get resultant R in i,j notation. R =R =R =R = 6.97 i + 1.22 j

11. Example 10 (Cont.): Find i,j for three vectors: A = 5 m,0 0 ; B = 2.1 m, 20 0 ; C = 0.5 m, 90 0. R = 7.08 m  = 9.93 0 N. of E. R = 6.97 i + 1.22 j 5. Determine R,  from x,y: R x = 6.97 m R  R y 1.22 m Diagram for finding R, 

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

13. A Graphical Understanding of the Components and of the Resultant is given below: 60 o 30 o R   Note: R x = A x + B x + C x AxAx B BxBx RxRx A C CxCx R y = A y + B y + C y 0 RyRy ByBy CyCy

14. Example 11 (Cont.) Using the Component Method to solve for the Resultant. 60 30 o R   AxAx B BxBx RxRx A C CxCx RyRy ByBy CyCy Write each vector in i,j notation. A x = 20 m, A y = 0 B x = -40 cos 60 o = -20 m B y = 40 sin 60 o = +34.6 m C x = -30 cos 30 o = -26 m C y = -30 sin 60 o = -15 m B = -20 i + 34.6 j C = -26 i - 15 j A = 20 i

15. Example 11 (Cont.) The Component Method 60 30 o R   AxAx B BxBx RxRx A C CxCx RyRy ByBy CyCy Add algebraically: A = 20 i B = -20 i + 34.6 j C = -26 i - 15 j R = -26 i + 19.6 j R -26 +19.6  R = (-26) 2 + (19.6) 2 = 32.6 m tan  = 19.6 -26  = 143 o

16. Example 11 (Cont.) Find the Resultant. 60 30 o R   AxAx B BxBx RxRx A C CxCx RyRy ByBy CyCy R = -26 i + 19.6 j R -26 +19.6  The Resultant Displacement of the bike is best given by its polar coordinates R and . R = 32.6 m;  = 143 0

17. Example 12. Find A + B + C for Vectors Shown below. A = 5 m, 90 0 B = 12 m, 0 0 C = 20 m, -35 0 A B R A x = 0; A y = +5 m B x = +12 m; B y = 0 C x = (20 m) cos 35 0 C y = -(20 m) sin -35 0 A = 0 i + 5.00 j A = 0 i + 5.00 j B = 12 i + 0 j B = 12 i + 0 j C = 16.4 i – 11.5 j C = 16.4 i – 11.5 j R =R =R =R = 28.4 i - 6.47 j C   CxCxCxCx CyCyCyCy

18. Example 12 (Continued). Find A + B + C A B C   R  R R x = 28.4 m R y = -6.47 m R = 29.1 m  = 12.8 0 S. of E.

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

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

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

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

23. 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, 30 0 ) RxRx RyRy R  Components of R: R x = R cos  R y = R sin 

24. Summary Continued: RxRx RyRy R  Resultant of Vectors:  Finding the resultant of two perpendicular vectors is like converting from polar (R,  ) to the rectangular (R x, R y ) coordinates.

25. 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 (R x,R y ).  Add vectors algebraically to get resultant in i,j notation. Then convert to (R 

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

27. Conclusion of Chapter - Vectors