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Vectors

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A set of slides created to teach Vectors to learners at Bishops Diocesan College in Cape Town.

A set of slides created to teach Vectors to learners at Bishops Diocesan College in Cape Town.

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Transcript

  • 1. Vectors & Scalars
    • Vectors
    • Quantities having both MAGNITUDE (size) and DIRECTION .
    • Eg. Displacement
    • Velocity
    • Acceleration
    • Force
    • For any vector both size and direction must be stated.
    • Scalars
    • Quantities having MAGNITUDE only.
    • Egs: Mass
    • Speed
    • Volume
  • 2.
    • Representing Vectors
    • All vector quantities can be represented by an arrow.
    • Magnitude = Length of LINE
    • The arrow is a straight line drawn to a suitable scale and the length of the arrow represents the magnitude of the vector while the arrow head represents the direction .
    • Scale: eg 1cm = 10m
    • The scale that is used must be chosen carefully in order for the whole drawing to fit on one page
    • the scale must be shown on the page.
    Direction ARROW
  • 3. Direction - Compass S NE NW SW SE Compass bearings can be used to indicate the direction of a vector. SSE WNW WNW = west of north west E W N
  • 4. Direction - Compass bearings Compass bearings can be used to indicate the direction of a vector. Measurements always from North Always measured clockwise. E W S N NE 45 o NW SW SE 90 o 135 o 180 o 225 o 270 o 315 o 360 o /0
  • 5. Direction - Compass S NE NW SW SE 45 o 45 o Compass bearings can be used to indicate the direction of a vector. ENE 10 o 280 o or 10 o N of W 67.5 o E W N
  • 6. Bearings Find the bearings for each of the vectors A – D. Two different ways for each vector. A B C D 30 o Clockwise measurements POSITIVE . Anticlockwise NEGATIVE. 30 o 30 o 30 o E W S N
  • 7. Bearings A = 30 o or 30 o E of N B = 120 o or 30 o S of E or E 30 o S C = 210 o or 30 o W of S or S 30 o W D = 300 o or 30 o N of W or W 30 o N 30 o A B C D 120 O 210 O 30 o E W S N
  • 8. Distance vs Displacement * A Start * B End Path traveled = DISTANCE Displacement = Straight line distance from starting point to finishing point. A B For rectilinear motion DISTANCE = DISPLACEMENT
    • Circular motion:
    • A - B displacement = diameter
    • distance = 1/ 2 circumference
    • A - A displacement = 0!
    • distance = circumference ( 2  r )
  • 9. Distance vs Displacement * A Start * B End Path traveled = DISTANCE
    • Distance cannot be less than displacement.
    • A “negative” displacement is a movement in the opposite direction to the one CHOSEN as POSITIVE.
    Right as + s 1 = 5m START s 1 s 2 = -1m s tot = +5 +(-1) = +4m
  • 10. Resultant Linear Displacements S 1 = 3km, 90º S 2 = 4km, 90 º 1. Same direction - A person walks 3km east and then 4km further east - find their ……………. displacement . S 1 = ………… S 2 = ………………… 2. Opposite direction - A person walks 3km east and then 4km West. (Take East as ……………….) R = ..………………….. Choose one direction (……….) as …………… S 1 = …….. S 2 = ……….. .: R = ……………………………………………………………. R = ……………………….. .: R = ……………………….. N
  • 11. Resultant Linear Displacements S 1 = 3km, 90º S 2 = 4km, 90 º 1. Same direction - A person walks 3km east and then 4km further east - find their resultant displacement . S 1 = 3km, 90º S 2 = -4km, (ie 270º) 2. Opposite direction - A person walks 3km east and then 4km West. (Take East as positive) R = 7km 90º R = -1km or 1km 270º Choose one direction (East) as positive S 1 = +3km S 2 = +4km .: R = S 1 + S 2 = +3 + 4 = +7km ie: 7km EAST R = S 1 + S 2 = +3 -4 = -1km .: R = 1km WEST N
  • 12. Example
    • b) Calculation :
    •  Choose ……………..
    •  West would therefore be
    • …………………… .
    • When we add the two vectors together
    •  R  ……………………
    • Vectors in the same or opposite direction
    • Find the RESULTANT of the following two vectors: 10m East and 6m West.
    • a) Construction: 5mm = 1m
    • 10m
  • 13. Example
    • b) Calculation :
    •  Choose East as positive
    •  West would therefore be
    • negative.
    • When we add the two vectors
    • above together we get
    •  R  +10m + -6m = +4m
    • Vectors in the same or opposite direction
    • Find the RESULTANT of the following two vectors: 10m East and 6m West.
    • a) Construction: 5mm = 1m
    • 10m
    Motion in a straight line is called rectilinear or linear motion.
  • 14. Resultant The RESULTANT (R ) of a number of vectors is the ……... ………….. that will have the ……………… as all the original vectors acting together. It stretches from the ………(tail) of the first vector to the …………… (…………..) of the last vector. (Tail to Head) Eg. Determine the resultant displacement of a person who walks 4km due east and then 3km north.
  • 15. Resultant 3km N 4km E N The RESULTANT (R ) of a number of vectors is the single vector that will have the same effect as all the original vectors acting together. It stretches from the beginning of the first vector to the end of the last vector. R= ? Eg. Determine the resultant displacement of a person who walks 4km due east and then 3km north.
  • 16. Resultant 3km N (3cm) 4km E (4 cm) N The resultant (R ) of a number of vectors is the single vector that will have the same effect as all the original vectors acting together. It stretches from the beginning of the first vector to the end of the last vector. R= 5cm Using Pythagoras 3 2 + 4 2 = 9 + 16 = 25 R 2 = 25 R = 5km 53.1 o E of N sinB = o/h = 3/5 B = sin -1 (0.6) B = 36.9 Bearing = 90 - 36.9 = 53.1 B
  • 17. “ Tail to Head” Method Boat rowed for 80m on a bearing of 30 o and then for 60m at 90 o . Find resultant displacement.
  • 18. “ Tail to Head” Method Boat rowed for 80m on a bearing of 30 o and then for 60m at 90 o . Find resultant displacement. 60m, 90 o 80m, 30 o 30 o 30 o 90 o N
  • 19. “ Tail to Head” Method Boat rowed for 80m on a bearing of 30 o and then for 60m at 90 o . Find resultant displacement. 60m, 90 o 80m, 30 o R = 122m, 55 o 30 o 30 o 90 o Construction N
  • 20. Parallelogram Method (Tail to Tail) Boat rowed for 80m on a bearing of 30 o and then for 60m at 90 o . Find resultant displacement. 60m, 90 o 80m, 30 o R = 122m, 55 o Calculation N
  • 21. Multiple Displacements POLYGON METHOD Find the resultant displacement that occurs after successive displacements of 3.4km @90 o , 2.9km@ 132 o , 4km@256 o , 3km@340 o.
  • 22. Multiple Displacements POLYGON METHOD Find the resultant displacement that occurs after successive displacements of 3.4km @90 o , 2.9km@ 132 o , 4km@256 o , 3km@340 o. R = 0.5km on a bearing of 90 o N
  • 23. Forces as a Vector Same direction Two forces of15 N at 90 o and 40 N at 90 o are applied to a box. Opposite direction Two forces of 100 N at 90 o and 40 N at270 o are applied to a box.
  • 24. Forces as a Vector Same direction 15 N 90 o and 40 N 90 o R = 15 + 40 = 55N 90 o 90 o (Right)is positive Opposite direction 100 N 90 o and 40 N 270 o R = -40 + 100 = +60N 90 o
  • 25. Forces at an angle Two forces at an angle. Same direction - angle 0 o What happens to the magnitude of the resultant if the angle between the forces increases?? N
  • 26. Forces at an angle Two forces at an angle of 40 o Resultant is now SHORTER N
  • 27. Forces at an angle As the angle between the forces increases the MAGNITUDE of the resultant DECREASES. N
  • 28. Forces at an angle As the angle between the forces increases the MAGNITUDE of the resultant DECREASES. N
  • 29. As the angle between the forces increases the MAGNITUDE of the resultant DECREASES. Forces at an angle N
  • 30. Forces at an angle As the angle between the forces increases the MAGNITUDE of the resultant DECREASES. N
  • 31. Forces at an angle As the angle between the forces increases the MAGNITUDE of the resultant DECREASES. The MINIMUM resultant is experienced when the forces are at 180 o . N
  • 32. Components of Vectors
    • Given Vector F
    • F can be expressed as the vector sum of two perpendicular vectors F x & F y
    F F x F y  y x
  • 33. Components of Vectors
    • Given Vector F
    • F can be expressed as the vector sum of two perpendicular vectors F x & F y
    F F x F y  F x = F cos  F y = F sin 
    • F x is the component of F in the x direction.
    • F y is the component of F in the y direction.
    y x
  • 34. Forces in equilibrium
    • 1. Consider 3 forces of 12N each, acting at 0  , 120  and 240  What would their resultant be?
    12N 12N 12N
    • The three vectors can be drawn as three sides of a ……………….. and there is ……. resultant.
    • The forces are ……………………. and therefore in ………………………..
    0 o
  • 35. Forces in equilibrium
    • 1. Consider 3 forces of 12N each, acting at 0  , 120  and 240 
    12N 12N 12N
    • The three vectors can be drawn as three sides of a triangle and there is no resultant.
    • The forces are balanced and therefore in equilibrium
    0 o
  • 36. The equilibrant
    • The equilibrant of a group of vectors is that single vector needed to produce equilibrium i.e. to balance the forces out.
    • Consider these two forces:
    • 12N at 0  and
    • 12N at 120 
    12N 12N The equilibrant is ……………… in magnitude but …………….. in direction to the resultant .
  • 37. The equilibriant
    • The equilibrant of a group of vectors is that single vector needed to produce equilibrium i.e. to balance the forces out.
    • Consider these two forces:
    • 12N at 0  and
    • 12N at 120 
    12N 12N Resultant Equilibriant The equilibriant is equal in magnitude but opposite in direction to the resultant .
  • 38. The equilibrant example
    • Two forces 10N at 90  and 8N at 120  act at a point on an object. Determine their resultant and equilibrant by accurate construction and measurement. Check your answer by calculation.
  • 39. The equilibrant example
    • Two forces 10N at 90  and 8N at 120  act at a point on an object. Determine their resultant and equilibrant by accurate construction and measurement. Check your answer by calculation.
    10N 8N R
  • 40. Eg - Hanging Weight
    • A weight of 30 N hangs on a rope from a ceiling. A horizontal force pulls it to the side. The angle between the rope and ceiling is 60 0 .
    • Determine: the tension in the rope (T) and the horizontal force.
    3kg 60 o S & M pg 16 & 17 F Gravity (weight)
  • 41. Eg - Hanging Weight
    • A weight of 30 N hangs on a rope fron a ceiling. A horizontal force pulls it to the side. The angle between the rope and ceiling is 60 0 .
    • Determine: the tension in the rope (T) and the horizontal force.
    3kg 60 o Force Diagram F g T F g =30N F T 60 o sin60 o = F g /T T = 30/sin60 o T = 34.6 N F h = S & M pg 16 & 17 F Gravity (weight)
  • 42. Inclined Plane
    • The system shown is in equilibrium.
    • What is the magnitude and direction of the friction force acting on the block?
    F f = ? F f = F g sin  m F g   F 90 F || N F f m 
  • 43. Inclined Plane
    • The system shown is in equilibrium. What is the magnitude and direction of the friction force acting on the block?
    250N 30 o
  • 44. Inclined Plane
    • The system shown is in equilibrium. What is the magnitude and direction of the friction force acting on the block?
    250N 30 o 30 o F f = F g sin 30 = 250(0.5) = 125N up the slope Sin30 = F f / F g F g
  • 45. Pulley system
    • The system shown is in equilibrium. What is the magnitude and direction of the friction force acting on the block?
    W 250N 30 o 10kg
  • 46. Pulley system
    • The system shown is in equilibrium. What is the magnitude and direction of the friction force acting on the block?
    W 10kg 250N 30 o 250N 250N 30 o 30 o 250N 100N F f = ? W 90 W || W ||
  • 47. Swimmer Problems
    • If the swimmer attempts to swim directly across the river what is his resultant velocity?
    • How long would it take to cross?
    • How far down the bank would he land?
    Current 1.5m.s -1 Width 30m Swimming speed 1.2m.s -1