CHAPTER THREE TWO DIMENSIONAL MOTION AND VECTORS

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.

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

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

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

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. 

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

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

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

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

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

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

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

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

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 = 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

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.

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 

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!

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  =  = o  = o

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 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

Example 8. Find the components of a 300- N force acting along the handle of a lawn- mower. The angle with the ground is 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

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, o ) Let 1 cm = 10 m