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VECTORS

VECTORS. THE MAGIC OF VECTOR MATH. 2 Types of Quantities. Scalars Just a Value This Value is called a Magnitude Vectors. &. Quantities that have  MAGNITUDE (size or value) AND DIRECTION. VECTORS. REPRESENTATION OF VECTOR QUANTITIES. VECTORS ARE REPRESENTED BY AN ARROW. tip.

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VECTORS

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  1. VECTORS THE MAGIC OF VECTOR MATH

  2. 2 Types of Quantities • Scalars • Just a Value • This Value is called a Magnitude • Vectors &

  3. Quantities that have  MAGNITUDE(size or value) AND DIRECTION VECTORS

  4. REPRESENTATION OF VECTOR QUANTITIES • VECTORS ARE REPRESENTED BY AN ARROW • tip • tail Click to Animate

  5. THE ARROW: LENGTH = • THE MAGNITUDE OR SIZE OF THE VECTOR THE ARROW’S DIRECTION = • IS THE DIRECTION OF THE VECTOR

  6. EXAMPLES OF VECTORS • FORCE (a push or a pull) • ELECTRIC/MAGNETIC FIELD STRENGTH • ACCELERATION • TORQUE – twist causing rotation • DISPLACEMENT – not distance • MOMENTUM – possessed by moving mass • VELOCITY – not speed

  7. EXAMPLES OF SCALARS • Mass • Time • Distance • Energy • + Everything else that’s not a vector….. These quantities have “NO DIRECTION”

  8. Answer questions 1 – 3 on the worksheet

  9. How to ADD VECTORS • Take care here • You Can NOT Add them like regular numbers (called Scalars)

  10. The SUM or RESULT of Adding 2 Vectors is called VECTOR ADDITION (THE TIP-TO-TAIL METHOD) FINDING THE RESULTANT Click to Animate A + B THE RESULTANT

  11. VECTOR ADDITION (THE TIP-TO-TAIL METHOD) FINDING THE RESULTANT Click to Animate A + B Yeilds A B or A B THE RESULTANT

  12. ADD VECTORS IN ANY ORDER (A+B = B+A) IF VECTORS ARE POINTING IN THE SAME DIRECTION  THIS IS REGULAR ALGABRAIC ADDITION REVIEWING VECTOR ADDITION Click to Animate

  13. POSITION THE TAIL OF ONE VECTOR TO THE TIP OF THE OTHER CONNECT FROM THE TAIL OF THE 1ST VECTOR TO THE TIP OF THE LAST THIS IS THE RESULTANT REVIEWING VECTOR ADDITION Click to Animate

  14. How About Vectors in Exactly Opposite DIRECTIONS ?

  15. ADDITION CONTINUED Click to Animate A B A B RESULTANT

  16. SUPPOSE THE VECTORS FORM A RIGHT ANGLE GRAPHICAL SOLUTIONS CAN ALWAYS BE USED BUT… HERE IS A MATHEMATICAL SOLUTION…. THIS SOLUTION USES THE PYTHAGOREAN THEORUM… C2 = A2 + B2 MORE VECTOR ADDITION

  17. ADDING VECTORS THAT ARE AT RIGHT ANGLES TO EACH OTHER… Click to Animate R = ?? lbs B = 3 lbs R2 = 16 + 9 = 25 R = 5 lbs A = 4 lbs BUT R = 5 lbs IS ONLY HALF AN ANSWER!! WHY????? R2 = A2 + B2 R2 = 42 + 32

  18. REMEMBER !!! Click to Animate • VECTORS HAVE 2 PARTS MAGNITUDE AND DIRECTION !!! HERE’S HOW TO FIND THE DIRECTION

  19. TRIG FUNCTIONS TO REMEMBER

  20. TRIG CALCULATIONS Click to Animate R = 5 lbs COS() = 4/5=0.8 B = 3 lbs SIN() = 3/5= 0.6  A = 4 lbs • USE YOUR CALCULATOR TO FIND THE ANGLE THAT HAS THESE VALUES OF SIN OR COS. Could also use TAN

  21. AT LAST THE ANGLE (THE VECTOR’S DIRECTION) Click to Animate SIN(X) = 0.6  ANGLE (X) = 37 DEGREES COS(X) = 0.8 ANGLE (X) = 37 DEGREES SO, THE OTHER HALF OF OUR ANSWER IS…..

  22. RESULTANT……. 5 lbs 37 degrees NORTH OF EAST • Not NORTHEAST • i.e. NE is 45 deg R = 5 lbs B = 3 lbs = 37 deg A = 4 lbs

  23. Answer questions 4 – 7 on the worksheet

  24. SUMMARY Click to Animate • A VECTOR IS A DIRECTED QUANTITY THAT HAS BOTH A MAGNITUDE AND DIRECTION • IF THE ANGLE BETWEEN THE VECTORS IS • 0 deg: algebraic addition (MAXIMUM ANS.) • 180 deg: algebraic “subtraction” (MINIMUM ANS.) • 90 deg: use Pythagorean Theorem to find magnitude and trig functions to find the angle

  25. THE END OF PART 1 !!!! RETURN TO BEGINNING • COULD YOU PASS A QUIZ ON THIS MATERIAL???? • NOW? • LATER, WITH STUDY? CONTINUE TO VECTOR MATH

  26. Vector Concepts used in Physics – Fancy Foot Work • Imagine you were asked to mark your starting place and walk 3 meters North, followed by two meters East. Could you answer the following: • How far did you walk? • Where are you relative to your original spot?

  27. Fancy Foot Work • How far did you walk? • This requires a MAGNITUDE ONLY • SCALAR QUANTITY called DISTANCE • 3m + 2m = 5m • Where are you relative to your original spot? • This requires both a MAGNITUDE & DIRECTION • VECTOR QUANTITY called DISPLACEMENT

  28. Answer question 8 on the worksheet

  29. 2m, E End 3m, N Displacement: Needs Magnitude & Direction Fancy Foot Work • Where are you relative to your original spot? • This requires both a MAGNITUDE & DIRECTION • VECTOR QUANTITY called DISPLACEMENT Start

  30. N 2m, E  W E 3m, N Displacement = s S Fancy Foot Work • Magnitude =? • The Length of the Hypotenuse • s2=(3m)2 + (2m)2 • s = • Direction =  East of North • Pick your Trig function • =33.7o, E of N

  31. 2m, E  = 3m, N Displacement = s = Fancy Foot Work NOW, measure the angle from the +X axis…… By convention, measure all angles Counterclockwise from the X Axis =33.7 90 – 33.7 = 56.3o

  32. Multiplying a Vector by a Scalar • When you multiply a vector by a scalar, it only affects the MAGNITUDE of the vector ** Not the direction** • Example:

  33. Answer questions 9-11 on the worksheet

  34. Vector Components • Component means “part” • A vector can be composed of many parts known as components • It’s best to break a vector down into TWO perpendicular components. WHY? • To use Right Triangle Trig 2 perpendicular components many components

  35. Vector Components • Introducing Vector V • Vector V’s X-Component is its Projection onto the X-axis • Vector V’s Y-Component is its Projection onto the Y-axis Sub Scripts in Action Now we have a Right Triangle Vy Vx

  36. Vector Components • Given this diagram, find V’s X & Y Components Vx= 5 4 Vy= =38.66o What’s the Magnitude of Vector V?

  37. Vx Vector Components • Now, knowing the magnitude of vector V, verify the V’s X & Y components using Trig Here's a check Vx=? Vy=? Vy 38.66o  

  38. Vector Components Important Stuff Gang • Golden Rules of Vector Components • 1. If you know the magnitude and direction of vector V to be (V,), then you can find Vx & Vy by • Vx=Vcos • Vy=Vsin • 2. If Vx & Vy are known, the magnitude of the vector can be found with Pythagorean Theorem: FIND THESE RULES IN YOUR PHYSICS DATA BOOKLET!!

  39. WHAT ARE THE X & Y COMPONENTS OF VECTOR ‘A’? THESE ARE The VECTOR COMPONENTS OF A Sin q=OPP/HYP Sin q=Ay/A Ay = Asinq Ax Cos q =ADJ/HYP Cos q = Ax/A Ax = Acosq A Ay Ay =Asinq q Ax =Acosq

  40. AN EXAMPLE • Suppose the magnitude of A = 5 and q = 37 deg. Find the VALUES of the X & Y components. COMPONENTS = 5 Ay 3 A Ay =Asinq 5sin 37 5(0.6) q =37 Ax =Acosq 5cos37 5(0.8) 4

  41. Answer question 12 on the worksheet

  42. MORE EXAMPLES

  43. QUESTION #1 A couple on vacation are about to go sight-seeing in a city where the city blocks are all squares. They start out at their hotel and tour the city by walking as follows: 1 block East; 2 blocks North; 3 blocks East; 3 blocks South; 2 blocks West;1 block South; 6 blocks East; 8 blocks North; 8 blocks West. WHAT IS THEIR DISPLACEMENT? (i.e., WHERE ARE THEY FROM THEIR HOTEL)?

  44. ANSWER #1: USING GRAPH PAPER 1 block East; 2 blocks North; 3 blocks East; 3 blocks South; 2 blocks West;1 block South; 6 blocks East; 8 blocks North; 8 blocks West. WHERE ARE THEY FROM THE HOTEL? THEY ARE 6 BLOCKS NORTH OF THE HOTEL H

  45. QUESTION #2 RIVER BOAT A river flows in the east-west direction with a current 6 mph eastward. A kayaker (who can paddle in still water at a maximum rate of 8 mph) wishes to cross the river in his boat to the North. If he points the bow of his boat directly across the river and paddles as hard as he can,what will be his resultant velocity?

  46. q = ARCTAN 6/8 = 37 deg. ANSWER #2 6 USE PYTHAGOREAN THEOREM 8 RESULTANT VELOCITY q R2 = 62 + 82 R2 = 36 + 64 R2 = 100 R = 10 R = 10 mph, 37 deg East of North or 10 mph, bearing 037 deg.

  47. QUESTION #3 BOAT RIVER Desired path to the South shore The kayaker wants to go directly across the river from the North shore to the South shore, again, paddling as fast as he can. At what angle should the kayaker point the bow of his boat so that he will travel directly across the river? What will be his resultant velocity?

  48. ANSWER #3... BOAT RIVER Desired path to the South shore q VEL.BOAT RESULTANT VELOCITY 8 mph R = arctan = arctan(opp/adj) =arc tan(6/5.3) = arctan 1.1 = 49 deg upstream q VEL.RIVER 82 = 62 + Vr2 Vr2 = 64 - 36 = 5.3 mph 6 mph

  49. 80 newtons 60 newtons QUESTION #4: A Two forces A and B of 80 and 60 newtons respectively, act concurrently(at the same point, at the same time) on point P. Calculate the resultant force. B P

  50. ANSWER #4: Pythagorean Theorem for 3-4-5 right triangle A R = 10 newtons RESULTANT 80 newtons = arctan (80/60) = 53 deg q q B P 60 newtons

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