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Physics is the Science of Measurement

Weight. Time. Physics is the Science of Measurement. Length. We begin with the measurement of length: its magnitude and its direction. B. A. Distance: A Scalar Quantity. Distance is the length of the actual path taken by an object. A scalar quantity:

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Physics is the Science of Measurement

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

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

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

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

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

  6. N 45o N W E W E 50o S S Identifying Direction Write the angles shown below by using references to east, south, west, north. 500 S of E Click to see the Answers . . . 450 W of N

  7. 90o 90o R 180o 180o 50o q 0o 0o 270o 270o Vectors and Polar Coordinates Polar coordinates (R,q) are an excellent way to express vectors. Consider the vector 40 m, 500 N of E, for example. 40 m R is the magnitude and qis the direction.

  8. 90o 180o 0o 60o 50o 60o 60o 3000 210o 270o 120o Vectors and Polar Coordinates Polar coordinates (R,q) are given for each of four possible quadrants: (R,q) = 40 m, 50o (R,q) = 40 m, 120o (R,q) = 40 m, 210o (R,q) = 40 m, 300o

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

  10. R y q x Trigonometry Review • Application of Trigonometry to Vectors Trigonometry y = R sin q x = R cos q R2 = x2 + y2

  11. 300 90 m Example 1:Find the height of a building if it casts a shadow 90 m long and the indicated angle is 30o. The height h is opposite 300 and the known adjacent side is 90 m. h h = (90 m) tan 30o h = 57.7 m

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

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

  14. N 400 m y = ? 30o E x = ? The x-component is: Rx = +346 m Example 2 (Cont.):A 400-m walk in a direction of 30o N of E. How far is the displacement east and how far north? Note: x is the side adjacent to angle 300 ADJ = HYP x Cos 300 x = R cosq x = (400 m)cos30o = +346 m, E

  15. N 400 m y = ? 30o E x = ? The y-component is: Ry = +200 m Example 2 (Cont.):A 400-m walk in a direction of 30o N of E. How far is the displacement east and how far north? Note: y is the side opposite to angle 300 OPP = HYP x Sin 300 y = R sinq y = (400 m) sin 30o = + 200 m, N

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

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

  18. x = R cosq y = R sinq Signs for Rectangular Coordinates 90o Second Quadrant: R is positive (+) 90o > q < 180o x = - ; y = + R + q 180o

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

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

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

  22. 40 lb 40 lb 30 lb 30 lb 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 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! 4 cm = 40 lb 3 cm = 30 lb

  23. q 40 lb f 40 lb 30 lb R = (40)2 + (30)2 = 50 lb R = x2 + y2 -30 40 30 lb tan f = Finding Resultant: (Cont.) Finding (R,q) from given (x,y) = (+40, -30) Rx Ry R q = 323.1o f = -36.9o

  24. 30 lb 30 lb R R R = 50 lb q 40 lb Ry q Ry f f q Rx Rx 40 lb 40 lb R 30 lb Rx Rx q 40 lb f Ry Ry R = 50 lb 30 lb R Four Quadrants: (Cont.) f = 36.9o; q = 36.9o; 143.1o; 216.9o; 323.1o

  25. 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

  26. R +40 m f -30 m Example 4: A woman walks 30 m, W; then 40 m, N. Write her displacement in i,j notation and in R,q notation. 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.

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

  28. 46 km f=? B 35 km R = ? A q = 1800 + 52.70 q = 232.70 Example 5: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.

  29. F = 300 N 32o 32o Fx 320 Fy Fy F Or in i,j notation: F = -(254 N)i- (159 N)j 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. Fx= -|(300 N) cos 320|= -254 N Fy= -|(300 N) sin 320|= -159 N

  30. 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).

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

  32. 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 7 (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.

  33. 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 7 (Cont.)Find resultant displacement. Now, We Find R,  R = 2.24 km  = 63.40 N or W

  34. 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

  35. 30 lb R q 40 lb Ry f q Rx 40 lb R 30 lb Rx Ry Significant Digits for Angles 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: q = 36.9o; 323.1o

  36. 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 Example8: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 ...)

  37. 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 Example8:Find R,q for the three vector displacements below: (A table may help.)

  38. 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 Example8(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 =

  39. R = 6.97 i + 1.22 j Diagram for finding R,q: R q Ry 1.22 m Rx= 6.97 m Example8(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.

  40. q Example9: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

  41. 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

  42. Cy By 30o B C R Ry A 60 f q Rx Ax Cx Bx Example9(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

  43. 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 Example9(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

  44. Cy By 30o B C R Ry A 60 f q Rx Ax Cx Bx R +19.6 f -26 Example9(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

  45. 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 10.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 =

  46. B 350 A C q q R R Example 10. (Continued).Find A + B + C Rx = 28.4 m Ry = -6.47 m R = 29.1 m q = 12.80 S. of E.

  47. 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

  48. 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.

  49. 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

  50. 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

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