Physics is the Science of Measurement Length Weight

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

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

Displacement—A Vector Quantity • Displacement is the straight-line separation of two points in a specified direction. D = 12 m, 20 o A B A vector quantity: Contains magnitude AND direction, a number, unit & angle. (12 m, 300; 8 km/h, N)

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

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

Identifying Direction Write the angles shown below by using references to east, south, west, north. N W 45 o E 50 o S N W E S 0 S of 50 Click to Esee the Answers 450 W. of. . N

Vectors and Polar Coordinates Polar coordinates (R, ) are an excellent way to express vectors. Consider the vector 40 m, 500 N of E, for example. 90 o 180 o 270 o 90 o 40 m 180 o 50 o 0 o 270 o R is the magnitude and is the direction.

Vectors and Polar Coordinates Polar coordinates (R, ) are given for each of four possible quadrants: 90 o (R, ) = 40 m, 50 o 120 o 210 o 180 o 60 o 50 o 60 o 3000 270 o 0 o (R, ) = 40 m, 120 o (R, ) = 40 m, 210 o (R, ) = 40 m, 300 o

Rectangular Coordinates y (-2, +3) (+3, +2) + (-1, -3) + x Reference is made to x and y axes, with + and - numbers to indicate position in space. Right, up = (+, +) - Left, down = (-, -) (+4, -3) (x, y) = (? , ? )

Trigonometry Review Application of Trigonometry to Vectors Trigonometry R y x = R cos q x y = R sin q R 2 = x 2 + y 2

Example 1: Find the height of a building if it casts a shadow 90 m long and the indicated angle is 30 o. The height h is opposite 300 and the known adjacent side is 90 m. h 300 90 m 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 = R cos R x y 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? N N R 400 m x y 30 o E y=? x=? The x-component (E) is ADJ: x = R cos The y-component (N) is OPP: y = R sin E

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? N Note: x is the side 400 m 30 o y=? x=? E x = (400 m) cos 30 o = +346 m, E adjacent to angle 300 ADJ = HYP x Cos 300 x = R cos The x-component is: Rx = +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? N Note: y is the side 400 m 30 o y=? x=? E opposite to angle 300 OPP = HYP x Sin 300 y = R sin y = (400 m) sin 30 o The y-component is: = + 200 m, N Ry = +200 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? N 400 m 30 o Rx = Ry = +200 m E The x- and ycomponents are each + in the first quadrant +346 m Solution: The person is displaced 346 m east and 200 m north of the original position.

Signs for Rectangular Coordinates 90 o First Quadrant: R is positive (+) R + + 0 o > < 90 o 0 o x = +; y = + x = R cos y = R sin

Signs for Rectangular Coordinates 90 o 180 o + R Second Quadrant: R is positive (+) 90 o > < 180 o x=-; y=+ x = R cos y = R sin

Signs for Rectangular Coordinates Third Quadrant: R is positive (+) 180 o > < 270 o x=- - R 270 o y=- x = R cos y = R sin

Signs for Rectangular Coordinates Fourth Quadrant: R is positive (+) + 360 o R 270 o > < 360 o x=+ y=- x = R cos y = R sin

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

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? Draw a rough sketch. Choose rough scale: Ex: 1 cm = 10 lb 40 lb 30 lb 40 lb Note: Force has direction just like cm =force 40 lb length does. We can 4 treat vectors just as we have 30 lb 3 cmlength = 30 lb vectors to find the resultant force. The procedure is the same!

Finding Resultant: (Cont. ) Finding (R, q) from given (x, y) = (+40, -30) 40 lb Rx R= tan = Ry R 30 lb x 2 + y 2 -30 40 R= 40 lb 30 lb (40)2 + (30)2 = 50 lb = -36. 9 o = 323. 1 o

Four Quadrants: (Cont. ) 30 lb Ry 40 lb R = 50 lb Rx 40 lb Rx 30 lb R Ry Rx 40 lb Rx Ry 30 lb R R R = 50 lb 40 lb Ry R 30 lb = 36. 9 o; 143. 1 o; 216. 9 o; 323. 1 o

Unit vector notation (i, j, k) y j k z Consider 3 D axes (x, y, z) i x 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

Example 4: A woman walks 30 m, W; then 40 m, N. Write her displacement in i, j notation and in R, notation. In i, j notation, we have: +40 m R -30 m R = R xi + R y 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.

Example 4 (Cont. ): Next we find her displacement in R, notation. +40 m R -30 m q = 1800 – 59. 10 = 126. 9 o R = 50 m (R, ) = (50 m, 126. 9 o)

Example 5: Town A is 35 km south and 46 km west of Town B. Find length and direction of highway between towns. 46 km R = -46 i – 35 j =? 35 km B R=? R = 57. 8 km A = 1800 + 52. 70 = 52. 70 S. of W. = 232. 70

Example 6. Find the components of a 300 -N force acting along the handle of a lawn-mower. The angle with the ground is 320. F = 300 N Fx 32 o Fx = -|(300 N) cos 320| = -254 N Fy = -|(300 N) sin 320| = -159 N Fy 320 F Fy Or in i, j notation: F = -(254 N)i - (159 N)j

Component Method 1. Start at origin. Draw each vector to scale with tip of 1 st to tail of 2 nd, tip of 2 nd to tail 3 rd, and so on for others. 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, ).

Example 7. 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. N 1. Start at origin. D 3 km, W Draw each vector to 2 km, S C B scale with tip of 1 st to 4 km, N tail of 2 nd, tip of 2 nd E to tail 3 rd, and so on A 2 km, E 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.

Example 7 (Cont. ) Find resultant displacement. 3. Write each vector in i, j notation: A = +2 i B= +4 j C = -3 i D= -2 j D 2 km, S N 3 km, W C B 4 km, N A 2 km, E E R = -1 i + 2 j 4. Add vectors A, B, C, D algebraically to get resultant in i, j notation. 1 km, west and 2 km north of origin. 5. Convert to R, notation See next page.

Example 7 (Cont. ) Find resultant displacement. Resultant Sum is: R = -1 i + 2 j Now, We Find R, D 2 km, S N 3 km, W C B 4 km, N A 2 km, E R = 2. 24 km R = 63. 40 N or W Rx = -1 km Ry= +2 km E

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

Significant Digits for Angles Since a tenth of a degree can often be significant, sometimes a fourth digit is needed. = 36. 9 o; 323. 1 o Rule: Write angles to the nearest tenth of a degree. See the two examples below: 30 lb R Ry Rx 40 lb Rx R 40 lb Ry 30 lb

Example 8: Find R, for the three vector displacements below: A = 5 m, 00 B = 2. 1 m, 200 C = 0. 5 m, 900 R A=5 m C= B m 0. 5 200 B = 2. 1 m 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. . . )

Example 8: Find R, for the three vector displacements below: (A table may help. ) For i, j notation find x, y components of each vector A, B, C. Vector A=5 m 00 C= R B A=5 m m 200 B = 2. 1 m X-component (i) +5 m Y-component (j) 0 B=2. 1 m 200 +(2. 1 m) cos 200 C=. 5 m 900 0 Rx = Ax+Bx+Cx +(2. 1 m) sin 200 + 0. 5 m Ry = Ay+By+Cy 0. 5

Example 8 (Cont. ): Find i, j for three vectors: A = 5 m, 00; B = 2. 1 m, 200; C = 0. 5 m, 900. X-component (i) Y-component (j) Ax = + 5. 00 m Ay = 0 Bx = +1. 97 m By = +0. 718 m Cx = 0 Cy = + 0. 50 m 4. Add vectors to get resultant R in i, j notation. A = 5. 00 i + 0 j B = 1. 97 i + 0. 718 j C= 0 i + 0. 50 j R = 6. 97 i + 1. 22 j

Example 8 (Cont. ): Find i, j for three vectors: A = 5 m, 00; B = 2. 1 m, 200; C = 0. 5 m, 900. R = 6. 97 i + 1. 22 j 5. Determine R, from x, y: Diagram for finding R, : R R = 7. 08 m Ry 1. 22 m Rx= 6. 97 m q = 9. 930 N. of E.

Example 9: 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? C = 30 m B= 40 m 30 o R Graphically, we use ruler and protractor to draw components, then measure the Resultant R, 60 o A = 20 m, E Let 1 cm = 10 m R = (32. 6 m, 143. 0 o)

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

Example 9 (Cont. ) Using the Component Method to solve for the Resultant. Cy B y Ry Write each vector in i, j notation. 30 o C R B Rx Cx 60 Ax = 20 m, Ay = 0 A Ax Bx Cx = -30 cos 30 o = -26 m Cy = -30 sin 60 o = -15 m A = 20 i Bx = -40 cos 60 o = -20 m By = 40 sin 60 o = +34. 6 m B = -20 i + 34. 6 j C = -26 i - 15 j

Example 9 (Cont. ) The Component Method Cy B y Ry 30 o C R B Rx B = -20 i + 34. 6 j 60 C = -26 i - 15 j A Ax Cx +19. 6 Add algebraically: A = 20 i R = -26 i + 19. 6 j Bx R -26 R= (-26)2 + (19. 6)2 = 32. 6 m tan = 19. 6 -26 = 143 o

Example 9 (Cont. ) Find the Resultant. R = -26 i + 19. 6 j Cy B 30 o y B C Ry R Rx Cx 60 A +19. 6 Ax R -26 Bx The Resultant Displacement of the bike is best given by its polar coordinates R and . R = 32. 6 m; = 1430

Example 10. Find A + B + C for Vectors Shown below. B Cx A = 5 m, 900 350 C y A B = 12 m, 00 C = 20 m, -35 R Ax = 0; Ay = +5 m Cx = (20 m) cos 350 A= 0 i + 5. 00 j B = 12 i + 0 j C = 16. 4 i – 11. 5 j Cy = -(20 m) sin -350 R = 28. 4 i - 6. 47 j Bx = +12 m; By = 0

Example 10. (Continued). Find A + B + C Rx = 28. 4 m B 350 A R C R Ry = -6. 47 m R = 29. 1 m q = 12. 80 S. of E.

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

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 R’ A A -B

Addition and 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 R=A+B R A A B R’ = A - B A R’ -B

Example 13. Given A = 2. 4 km, N and B = 7. 8 km, N: find A – B and B – A. A – B; B-A A-B +A -B R A B 2. 43 N 7. 74 N B-A +B -A R (2. 43 N – 7. 74 S) (7. 74 N – 2. 43 S) 5. 31 km, S 5. 31 km, N

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) Components of R: Rx = R cos q Ry = R sin q R Rx Ry

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

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).

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 R’ A A -B