1. 15 Mathematical Fundamentals Need working knowledge of algebra and basic trigonometry if you don’t have this then you must see me immediately!

2. 16 Algebra Review Exponents - Square Roots 5 2 exponent 5 * 5 = 25 2 3 = 2 * 2 * 2 = 8 25= 25 1/2 = 5

3. 17 Order of Operations Solve the following problem (12 + * 3)2-1 + 3 * 2 - (8/4) - 5 2 - 6/2 = ??? 2 3

4. 18

5. 19 Order of Operations (1)parentheses, brackets, and braces (2) exponents, square roots (3) multiplication and division (4) addition and subtraction

6. 20 Order of Operations Problem SOLUTION (12 + * 3)2-1 + 3 * 2 - (8/4) - 5 2 - 6/2 = ??? 2 3 1. parentheses (12+ *3) 1a. * 3 = 2 1b. 12 + 2 = 14 1c. 8/4 = 2 2 3 2 3

7. 21 Order of Operations Problem SOLUTION (12 + * 3)2-1 + 3 * 2 - (8/4) - 5 2 - 6/2 = ??? 2 3 2. exponents 5 2 = 25 3. multiplication & division (12 + *3)2 = 14*2 = 28 NOTE: 14 was calculated in steps 1a and 1b. 6/2 = 33*2 = 6 2 3

8. 22 Order of Operations Problem SOLUTION (12 + * 3)2-1 + 3 * 2 - (8/4) - 5 2 - 6/2 = ??? 2 3 Substitute into equation 28 -1 + 6 - 2 - 25 - 3 = 3

9. 23 Trigonometry field of mathematics focusing on relationships between sides of and the angles within a right triangle

10. 24 Trigonometry Review  c a b a = “opposite” side b = “adjacent” side c = “hypotenuse”  = angle

11. 25 SOHCAHTOA  c a b 4 Basic Relationships 1. a 2 + b 2 = c 2 (Pythagorean Theorem) 2. sin  = opp/hyp = a/c 3. cos  = adj/hyp = b/c 4. tan  = opp/adj = a/b a = “vertical component” b = “horizontal component” c = “resultant”

12. 26 Two types of TRIG problems Type A Type B GivenSolve For c &  a & b a & b c & 

13. 27 TYPE A Problem v = 10 m/s 40 o b a Given: c = 10 m/s  = 40 degrees Find:a and b sin 40 o = a 10 m/s 10 m/s * sin = a 10 m/s 40 o cos 40 o = b 10 m/s 10 m/s * cos = b 10 m/s 40 o b = 10 m/s * cos 40 = 7.66 m/s a = 10 m/s * sin 40 = 6.43 m/s o o

14. 28 Type B Problem 100 lb 400 lb c  Given: a = 400 lb, b = 100 lb Find: c and  a 2 + b 2 = c 2 (400 lb) 2 + (100 lb) 2 = c 2 160000 lb 2 + 10000 lb 2 = c 2 170000 lb 2 = c 2 c = 412.3 lb a tan  = b 400 lb tan  = 100 lb tan  = 4 tan -1 (tan  ) = tan -1 (4)  = 76.0 o

15. 29 Inverse Trig Functions If sin is a trig function then sin -1 is an inverse trig function :inverse trig functions simply “undo” trig functions

16. 30 SOHCAHTOA SOH * Sine = Opposite/Hypotenuse CAH * Cosine = Adjacent/Hypotenuse TOA * Tangent = Opposite/Adjacent

17. 31 25 20 o b a Calculate the vertical (a) and horizontal sides of this right triangle.

18. 32 25 20 o b a sin 20 = a 25 o cos 20 = b 25 o a = 25 (sin 20) a = 8.55 b = 25 (cos 20) b = 23.49

19. 33 10 15 c  Solve for the length of the hypotenuse (c) and the angle, .

20. 34 10 15 c  c = 15 2 + 10 2 c = 325 c = 18.03 tan  = 15 10  = tan -1 (1.5)  = 56.3 o

21. 35 UNITS Use the SI system –AKA Metric System –4 basic units length--meter mass--kilogram time--second temperature--degree Kelvin (Celsius)

22. Radio Flyer 50 lbs 45 o Billy pulls on his new wagon with 50 lbs of force at an angle of 45. How much of this resultant force is actually working to pull the wagon horizontally? Vector Resolution Example

23. FxFx FyFy F F = F x + F y 45 o F = magnitude of F = 50 lbs cos 45 = sin 45 = FxFx FyFy F F o o

24. FxFx FyFy F F = F x + F y 45 o = cos 45 = sin 45 FyFy FxFx F F o o FxFx FyFy = 50 lbs (cos 45 ) = 50 lbs * 0.707 = 35.4 lbs = 50 lbs (sin 45 ) = 50 lbs * 0.707 = 35.4 lbs o o

25. Radio Flyer 50 lbs 45 o Sometimes the magnitude of a force is written more simply as F x = 35.4 lbs F y = 35.4 lbs Only the force acting in the x-direction acts to move the wagon forward

26. 40 Vector Decomposition aka Vector Resolution Any vector can be expressed as a pair of two component vectors these vectors 1) must be perpendicular to each other 2) are usually horizontal and vertical

27. 41 Given the polar notation of a vector, decompose it into vertical and horizontal components (Cartesian coordinates). s x = |S|cos  = 10(.766) = 7.66 m s y = |S|sin  = 10(.643) = 6.43 m  = 40 o S = 10 m 10cos(40) 10sin(40) Vector Decomposition x y

28. Vector Composition (aka Vector Addition) to add 2 vectors must consider both magnitude and direction the sum of 2 or more vectors is known as a resultant vector if the vectors have the same direction then you may add the magnitudes directly + =

29. vectors are in opposite direction –resultant vector points in direction of longer vector –size of resultant vector is the difference between the component vectors + =

30. vectors are pointed in different, non- parallel, direction graphical solution - TIP-TO-TAIL method +

31. TIP-TO-TAIL method place the tail of the 2nd vector at the tip of the 1st vector connect the tail of the 1st vector to the 2nd vector + = resultant vector Resultant vector is the diagonal of the resulting parallelogram

32. TIP-TO-TAIL method is the preferred method when adding more than 2 vectors –include more vectors by attaching their tail to the open tip in the diagram + ++ +

33. + ++ +

34. Vector Example Graphically compute the resultant force acting on the femoral head. Two Forces Acting on the Hip muscle body weight W

35. resultant force acting on the femoral head FmFm W W R R = F m +W

36. 50 Vectors can be added by placing the tail of each vector at the tip of the previous one. The sum of all of these vectors is called the resultant vector. It connects the tail of the first vector to the head of the last vector. resultant Vector Addition

37. 51 Finding the horizontal and vertical components of each vector makes it easy to find the resultant. Vector Addition

38. 52 Simply add all of the vertical lines for the vertical component and add all of the horizontal lines for the horizontal component. Be sure to pay attention to the sign of each of the lines. resultant Vector Addition

39. 53 Use the following formulas to convert the coordinates into polar notation:  = arctan |S||S|  Vector Addition SxSx SySy

40. 54 y x S 2 = 3m, 165 o S 1 = 6m, 40 o

41. 55 y x S 2 = 3m, 165 o S 1 = 6m, 40 o

42. 56 Sx 1 = |S 1 |cos  1 = 6(.799) = 4.60 m Sy 1 = |S 1 |sin  1 = 6(.643) = 3.86 m Sx 2 = |S 2 |cos  2 = 3(-.966) = -2.90 m Sy 2 = |S 2 |sin  2 = 3(.259) =.78 m Sx = 4.60 - 2.90 = 1.70 m Sy = 3.86 +.78 = 4.64 m y x S 2 = 3m, 165 o S 1 = 6m, 40 o

43. 57 |S| = = 4.94 m  = arctan = 69.9 o Polar Notation