2. CHAPTER THREE TWO DIMENSIONAL MOTION AND VECTORS

3. Objectives: After completing this module, you should be able to: Demonstrate that you meet mathematics expectations: unit analysis, algebra, scientific notation, and right-triangle trigonometry. Define and give examples of scalar and vector quantities. Determine the components of a given vector. Find the resultant of two or more vectors. Demonstrate that you meet mathematics expectations: unit analysis, algebra, scientific notation, and right-triangle trigonometry. Define and give examples of scalar and vector quantities. Determine the components of a given vector. Find the resultant of two or more vectors.

4. Expectations (Continued) You must have mastered right-triangle trigonometry. y x R  y = R sin  x = R cos  R 2 = x 2 + y 2

5. Physics is the Science of Measurement We begin with the measurement of length: its magnitude and its direction. Length Weight Time

6. Distance: A Scalar Quantity A scalar quantity: Contains magnitude only and consists of a number and a unit. (20 m, 40 mi/h, 10 gal) A B  Distance is the length of the actual path taken by an object. s = 20 m

7. Displacement—A Vector Quantity A vector quantity: Contains magnitude AND direction, a number, unit & angle. (12 m, 30 0 ; 8 km/h, N) A B D = 12 m, 20 o Displacement is the straight-line separation of two points in a specified direction.Displacement is the straight-line separation of two points in a specified direction. 

8. Distance and Displacement Net displacement: 4 m,E 6 m,W D What is the distance traveled? 10 m !! D = 2 m, W Displacement is the x or y coordinate of position. Consider a car that travels 4 m, E then 6 m, W.Displacement is the x or y coordinate of position. Consider a car that travels 4 m, E then 6 m, W. x= +4 x = +4 x= -2 x = -2

9. Identifying Direction A common way of identifying direction is by reference to East, North, West, and South. (Locate points below.) 40 m, 50 o N of E EW S N 40 m, 60 o N of W 40 m, 60 o W of S 40 m, 60 o S of E Length = 40 m 50 o 60 o

10. Identifying Direction Write the angles shown below by using references to east, south, west, north. EW S N 45 o EW N 50 o S 50 0 S of E 45 0 W of N

11. Example 1: Find the height of a building if it casts a shadow 90 m long and the indicated angle is 30 o. 90 m 30 0 The height h is opposite 30 0 and the known adjacent side is 90 m. h h = (90 m) tan 30 o h = 57.7 m

12. Finding Components of Vectors A component is the effect of a vector along other directions. The x and y components of the vector (R,  are illustrated below. x y R  x = R cos  y = R sin  Finding components: Polar to Rectangular Conversions

13. Example 2: A person walks 400 m in a direction of 30 o N of E. How far is the displacement east and how far north? x y R  x = ? y = ? 400 m   E N The y-component (N) is OPP: The x-component (E) is ADJ: x = R cos  y = R sin  E N

14. Example 2: A person walks 400 m in a direction of 30 o N of E. How far is the displacement east and how far north? x y R  x = ? y = ? 400 m   E N The y-component (N) is OPP: The x-component (E) is ADJ: x = R cos  y = R sin  E N

15. Example 2 (Cont.): A 400-m walk in a direction of 30 o N of E. How far is the displacement east and how far north? x = R cos  x = (400 m) cos 30 o = +346 m, E x = ? y = ? 400 m   E N Note: x is the side adjacent to angle 30 0 ADJ = HYP x Cos 30 0 The x-component is: R x = +346 m

16. Example 2 (Cont.): A 400-m walk in a direction of 30 o N of E. How far is the displacement east and how far north? y = R sin  y = (400 m) sin 30 o = + 200 m, N x = ? y = ? 400 m   E N OPP = HYP x Sin 30 0 The y-component is: R y = +200 m Note: y is the side opposite to angle 30 0

17. Example 2 (Cont.): A 400-m walk in a direction of 30 o N of E. How far is the displacement east and how far north? R x = +346 m R y = +200 m 400 m   E N The x- and y- components are each + in the first quadrant Solution: The person is displaced 346 m east and 200 m north of the original position.

18. Resultant of Perpendicular Vectors Finding resultant of two perpendicular vectors is like changing from rectangular to polar coord. R is always positive;  is from + x axis x y R 

19. Example 3: A 30-lb southward force and a 40-lb eastward force act on a donkey at the same time. What is the NET or resultant force on the donkey? 30 lb 40 lb Draw a rough sketch. Choose rough scale: Ex: 1 cm = 10 lb 4 cm = 40 lb 3 cm = 30 lb 40 lb 30 lb Note: Force has direction just like length does. We can treat force vectors just as we have length vectors to find the resultant force. The procedure is the same!

20. Finding Resultant: (Cont.) 40 lb 30 lb 40 lb 30 lb Finding (R,  ) from given (x,y) = (+40, -30) R   RyRy RxRx R = x 2 + y 2 R = (40) 2 + (30) 2 = 50 lb tan  = -30 40  = -36.9 o  = 323.1 o

21. Example 7. Find the components of the 240-N force exerted by the boy on the girl if his arm makes an angle of 28 0 with the ground. 28 0 F = 240 N F FyFyFyFy FxFxFxFx FyFyFyFy F x = -|(240 N) cos 28 0 | = -212 N F y = +|(240 N) sin 28 0 | = +113 N Or in i,j notation: F = -(212 N)i + (113 N)j

22. Example 8. Find the components of a 300- N force acting along the handle of a lawn- mower. The angle with the ground is 32 0. 32 0 F = 300 N F FyFyFyFy FxFxFxFx FyFyFyFy F x = -|(300 N) cos 32 0 | = -254 N F y = -|(300 N) sin 32 0 | = -159 N 32 o Or in i,j notation: F = -(254 N)i - (159 N)j

23. 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