Mathematical Fundamentals

# Mathematical Fundamentals

## Mathematical Fundamentals

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##### Presentation Transcript

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

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

3. 2 3 Order of Operations • Solve the following problem (12 + * 3)2-1 + 3 * 2 - (8/4) - 52 - 6/2 = ???

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

5. 2 (12 + * 3)2-1 + 3 * 2 - (8/4) - 52 - 6/2 = ??? 3 1. parentheses (12+ *3) 1a. * 3 = 2 1b. 12 + 2 = 14 1c. 8/4 = 2 2 3 2 3 Order of Operations ProblemSOLUTION

6. 2 (12 + * 3)2-1 + 3 * 2 - (8/4) - 52 - 6/2 = ??? 3 2 3 Order of Operations ProblemSOLUTION 2. exponents 52 = 25 3. multiplication & division (12 + *3)2 = 14*2 = 28 NOTE: 14 was calculated in steps 1a and 1b. 6/2 = 3 3*2 = 6

7. 2 (12 + * 3)2-1 + 3 * 2 - (8/4) - 52 - 6/2 = ??? 3 Order of Operations ProblemSOLUTION Substitute into equation 28 -1 + 6 - 2 - 25 - 3 = 3

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

9. c a q b Trigonometry Review a = “opposite” side b = “adjacent” side c = “hypotenuse” q = angle

10. c a q b SOHCAHTOA 4 Basic Relationships 1. a2 + b2 = c2 (Pythagorean Theorem) 2. sin q = opp/hyp = a/c 3. cos q = adj/hyp = b/c 4. tan q = opp/adj = a/b a = “vertical component” b = “horizontal component” c = “resultant”

11. Two types of TRIG problems Given Solve For c & q a & b a & b c & q Type A Type B

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

13. a tan q = b 400 lb tan q = 100 lb Type B Problem Given: a = 400 lb, b = 100 lb Find: c and q c 400 lb q 100 lb a2 + b2 = c2 (400 lb)2 + (100 lb)2 = c2 160000 lb2 + 10000 lb2 = c2 170000 lb2 = c2 c = 412.3 lb tan q = 4 tan-1 (tan q) = tan-1(4) o q = 76.0

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

15. SOHCAHTOA • SOH • Sine = Opposite/Hypotenuse • CAH • Cosine = Adjacent/Hypotenuse • TOA • Tangent = Opposite/Adjacent

16. 25 a o 20 b Calculate the vertical (a) and horizontal sides of this right triangle.

17. 25 a o 20 b a o sin 20 = 25 b o cos 20 = 25 a = 25 (sin 20) a = 8.55 b = 25 (cos 20) b = 23.49

18. c 15 q 10 Solve for the length of the hypotenuse (c) and the angle, q.

19. c 15 q 15 10 tan q = 10 q = tan-1 (1.5) q = 56.3 o c = 152 + 102 c = 325 c = 18.03

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

21. 50 lbs o 45 Radio Flyer Vector Resolution Example 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?

22. Fx F Fy F o 45 F = Fx + Fy F = magnitude of F = 50 lbs o cos 45 = sin 45 = F o Fy Fx

23. Fx F F Fy o 45 Fx Fy F = Fx + Fy o = cos 45 = sin 45 o F Fy Fx o = 50 lbs (cos 45 ) = 50 lbs * 0.707 = 35.4 lbs o = 50 lbs (sin 45 ) = 50 lbs * 0.707 = 35.4 lbs

24. 50 lbs o 45 Radio Flyer Sometimes the magnitude of a force is written more simply as Fx = 35.4 lbs Fy = 35.4 lbs Only the force acting in the x-direction acts to move the wagon forward

25. 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

26. Vector Decomposition Given the polar notation of a vector, decompose it into vertical and horizontal components (Cartesian coordinates). • sx = |S|cosq = 10(.766) = 7.66 m • sy = |S|sinq = 10(.643) = 6.43 m S = 10 m 10sin(40) q = 40o x 10cos(40) y

27. 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 = +

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

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

30. Resultant vector is the diagonal of the resulting parallelogram resultant vector + = • 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

31. + + + + • 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

32. + + + +

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

34. R = Fm+W W Fm R W resultant force acting on the femoral head

35. resultant Vector Addition • 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.

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

37. resultant Vector Addition • 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.

38. Vector Addition • Use the following formulas to convert the coordinates into polar notation: • q = arctan |S| Sy q Sx

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

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

41. S2 = 3m, 165o Sx1 = |S1|cosq1 = 6(.799) = 4.60 m Sy1 = |S1|sinq1 = 6(.643) = 3.86 m Sx2 = |S2|cosq2 = 3(-.966) = -2.90 m Sy2 = |S2|sinq2 = 3(.259) = .78 m Sx = 4.60 - 2.90 = 1.70 m Sy = 3.86 + .78 = 4.64 m y S1 = 6m, 40o x

42. Polar Notation • |S| = = 4.94 m • q = arctan = 69.9o