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Vectors and Scalars

Vectors and Scalars. Vectors & Scalars. Vectors are the physical quantities having ; 1- Magnitude 2- Direction 3- Should obey the parallelogram law of addition

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Vectors and Scalars

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  1. Vectors and Scalars

  2. Vectors & Scalars Vectors are the physical quantities having ; 1- Magnitude 2- Direction 3- Should obey the parallelogram law of addition For Ex- Force ,Acceleration , weight are vectors because to define them direction and magnitude both are needed . They also follow the parallelogram law of addition. Scalar quantity :- Those physical quantities which requires only magnitude to explain it . Mass, distance , work , energy are scalar quantities ,because to define them magnitude is sufficient.

  3. Distance is scalar quantity whereas displacement is vector quantity Displacement is vector quantity :- Aero plane and ship required a fixed direction to go ahead as there are no roads in the sky or in ocean. So distance cover by then is known as displacement , which is vector quantity. Distance is scalar quantity :- In roads Bus generally moves according to zig-zag path ,because the Bus does not follow a fixed direction so distance cover by it is treated as scalar.

  4. C C C C Click to start Mass and Weight Mass- It is represented by m. It is the amount of matter which a body possess. The more mass an object has, the more force is needed to get it moving or slow it. It is scalar quantity. Weight -: It is the force by which earth attracts body towards its centre. It is represented by mg, where g ( for earth g is about 9.8 m/sec2) is the gravitational acceleration. It is vector quantity. Direction of Earth Pull Earth

  5. Types of vectors Unit Vector – It is the vector having magnitude one ‘1’. If displacement is 5m east . It means that its unit vector is 1m east. If a = a1î+a2Ĵ+a3kˆ then unit vector of it is represented as â (To distinguish this vector from others it is represented as â.), where â = (a1î+a2Ĵ+a3k ˆ )/(a12+a22+a32) If b = 2î+3Ĵ+4kˆ then bˆ = (2î+3Ĵ+4kˆ)/  (22+32+42) = (2î+3Ĵ+4kˆ)/  39 Zero Vector – It is the vector having magnitude zero. It may have any direction. It may be represented as AA, BB or CC etc.

  6. Parallel vectors:- If the vectors are parallel to each other then they are called parallel vectors. Here vector l , m ,n are parallel to each other. Vector product of two parallel vectors is zero, i.e. is l x m =0 m = l and n = l l m n Perpendicular vectors :- If the vectors are perpendicular to each other than they are called perpendicular vectors. Scalar product of two perpendicular vectors ( a & b) is zero i.e. a.b =0 Here vectors a & b are perpendicular to each other. l m

  7. Parallelogram law of addition R B C Q α α O D P A Click to start QSinα Q QCosα In OBD R2 = (P+QCosα)2 + Q2Sin2α R2 = (P2+Q2Cos2α + 2PQCosα) + Q2Sin2α = (P2+Q2Cos2α + 2PQCosα) + Q2Sin2α = P2+Q2 ( Cos2α +Sin2α)+2PQCosα = P2+Q2 +2PQCosα

  8. R= 5 i2= 3Amp I1 = 3Amp • On the other part current despite of having direction and magnitude does not obey the parallelogram law of addition In the above figure the resultant current should be 5 Amp, but it is actually 7 amp(3+4). It shows that current is a scalar quantity.

  9. Addition of many vectors α3 c c c c α2 d d d d R R b b b b α1 a a a a See adding of vectors See angle between different vectors b c α2 α1 α3 Here R = a + b+ c+ d Click to start Given vectors Here we are interested to get a+b+c+d Step wise way of addition

  10. RSinα R α RCosα P Mg Click to start How a Person moves? When a person push the earth backward, the earth provides the force of reaction “R” to person. R can be divided into two components, RSinα and RCosα . RSinα helps to make the weight of person light and RCosα helps to push a person forward. R is the reaction force given by earth. Here apparent weight of the body = Mg - Sinα Pushing the earth backward

  11. F2 F3 4 3 2 1 F1 F5 F4 R  If many forces are working at a point, then finding of resultant Here these forces can be break in their perpendicular components X & Y. X = F1 Cos0 + F2 Cos1 + F3Cos 2 +F4Cos 3 + F5Cos 4 Y = F1 Sin0 + F2 Sin1 + +F3Sin 2+ F4Sin 3 + F5Sin 4 So the resultant vector R =  X2 + Y2 I f resultant R makes an angle with X axis then tan = Y/X

  12. The another way of addition may be 3-2 F4 F3 2-1 4 F5 4-3 F2 F3 3 F2 R = F1+F2+F3+F4+F5 2 1 1 F1 F1 F5 F4

  13. Click to start Click to start Force is a vector quantity because it is either pull or push. Pull or push have direction also in addition to magnitude, so it is vector quantity. It is pulling. Direction of pulling is shown. It is pushing. Direction of force is shown in figure.

  14. Pulling of Roller is easier than pushing Click to start F Sinα F (Pulling) α F Cosα Pulling This force can be divided into two components. Here net weight (downward force on roller )of roller after applying force= Mg – FSinα The net horizontal pulling force = FCosα It shows that here the apparent weight of roller decreases, So it is easier to pull a roller instead of pushing. Roller Mg

  15. Click to start F Sinα F (Pushing) α F Cosα Pushing Here net weight (downward force on roller )of roller = Mg + FSinα The net horizontal pulling force = FCosα This force can be divided into two components. It shows that here the apparent weight of roller increases, So it is easier to pull a roller instead of pushing. Roller Note- In pushing or pulling angle ‘α’ remains same. Mg

  16. Path of Projectile v vSin1 1 vCos1 H= It is the maximum height attained by the projectile. u uSin  uCos C uCos = vCos1 ( Here vertical component becomes zero.) B vCos2 2 A v vSin2 (Range ‘R’:- It is the maximum distance cover by the projectile ) uCos = vCos1 = vCos2(Horizontal components of velocity always remains same.) vSin1 = uSint1 -1/2gt12 Relation between u & v at point A & B

  17. u u Position of particle after time ‘t’ uSin uSinα y   x uCos uCosα When a particle/body is thrown in the sky it’s path becomes parabolic. x= uCos.t( Horizontal component of velocity (uCos) is responsible for covering horizontal distance ‘x’.) y= uSInt – ½ gt2( Vertical component of velocity (uSIn) is responsible for covering vertical distance ‘y’.) Or y = uSin(x/uCos ) – 1/2g(x/uCos)2 OrY = x tan - (gSec2/2u2)x2 This equation is of the form Y = Ax + BX2 Which is the equation of parabola. Hence the path of projectile is parabolic. (x,y)

  18. Flight Time--Let a body takes time ‘t’ to reach in the highest point C. At C the vertical velocity of the body will be zero. So v = uSin - gt 0 = uSin - gt or t= uSin/g So total time taken T =2t = 2uSin/g RangeR = horizontal velocity x total flight time Range R = uCos x T (horizontal component of velocity is responsible for the horizontal distance covered by the body.) =uCos x 2uSin/g = u2 2SinCos/g = u2 Sin2/g Max value of Sin2 =1 So 2= 900 or  =450 Therefore max R = u2/2g Vertical Height H At point C the vertical component of velocity =0 ,So by v2= u2 - 2gH 0= u2Sin2 - 2gH or H= u2Sin2/2g Max value of Sin =1 so  = 900 Therefore maximum vertical height which can be attained H= u2/2g

  20. 450 R= longest range When  = 450 , then body covers the longest range. A hammer thrower should throw the hammer at the angle of 450 in order to cover maximum distance.

  21. 700

  22. When  = 900 , then body get the greatest height. High jumper should take lift at an angle of 900 in order to cover maximum vertical distance.

  23. Vector represents in 3 dimension Click to start Y 2i + 3j + 4k X 900 900 900 i j k Z If a vector is 2i + 3j +4k then it means that component of vector which is in the direction of X ,Y and Z axis are 2,3 and 4 respectively.

  24. Vector Products ^ ^ ^ ^ ^ n n n n n b b   a a Direction of aXb is perpendicular to both a & b. It is shown by . Area covered by a & b can be calculated as aXb b a If two vectors are parallel to each other then  = 0 , so a x b = absin0 = 0 . Click to start Vectors are multiplied in two ways 1-Vector product 2- Scalar Product 1-Vector product :- If a and b are two vectors then axb = abSin axb Force is the vector quantity, displacement is the vector quantity , because their product which give rise to moment is also a vector quantity , so here we take vector product. T(Torque) = Fxd = FdSin

  25. Scalar Product a. b = abCos b  a If two vectors are Perpendicular to each other then  = 90 , so a.b = abCos90 = 0 . a b Click to start 2-Scalar product :- If a and b are two vectors then a.b = abCos Here products of two vectors give rise to a scalar quantity. Force is the vector quantity, displacement is the vector quantity ,but their product which give rise to work is a scalar quantity. W = F.d = FdCos

  26. Ĵ Î kˆ Example of vector product:- If a = (2î+3Ĵ+4kˆ) and b = (4î+5Ĵ+6kˆ) then axb = (2î+3Ĵ+4kˆ)x(4î+5Ĵ+6kˆ) = 8îx î+ 10 îxj + 12 Îxkˆ + 12 Ĵxi + 15Ĵx Ĵ +18Ĵxkˆ +16 kˆx î +20kˆx Ĵ +24kˆxkˆ = 0 +10kˆ +12(-Ĵ) +12(-kˆ)+0 +18î+16(- Ĵ)+20(-î)+0 = 10kˆ -12 Ĵ – 12k +18 î -16 Ĵ – 20 î = -2 î – 28 Ĵ -2kˆ Important to know-: Î x Î = I Î I I Î I Sin00 kˆ = (1)(1)(0) kˆ = 0 (Angle between same vectors is zero.) Similarly Ĵ. Ĵ = kˆ.kˆ = 0 Îxj = I Î II Ĵ I Sin 900 kˆ= (1)(1)(1) kˆ = kˆ (Î, Ĵ ,kˆ are the perpendicular vectors to each other.) Similarly Ĵx kˆ = Î kˆxÎ = Ĵ kˆxĴ = - Î Î xkˆ = - Ĵ , Ĵ x Î = -kˆ

  27. Example of scalar product:- If a = (2î+3Ĵ+4kˆ) and b = (4î+5Ĵ+6kˆ) a.b = (2î+3Ĵ+4kˆ). (4î+5Ĵ+6kˆ) = 8î. î+ 10 î.j + 12 Î.kˆ + 12 Ĵ.i + 15Ĵ. Ĵ +18Ĵ.kˆ +16 kˆ. î +20kˆ. Ĵ +24kˆ.kˆ = 8 +0 +0 +0+15 +0+0+0+24 = 47 Important to know-: Î . Î = I Î II Î I Cos0 = (1)(1)(1) = 1 (Angle between same vectors is zero.) Similarly Ĵ. Ĵ = kˆ.kˆ = 1 Î. j = I Î II Ĵ I Cos 900 = 0 (Î, Ĵ ,kˆ are the perpendicular vectors to each other.) Similarly Ĵ. kˆ = kˆ.Î = 0

  28. (x2 < x1) & (t2 > t1) Gradient = x2-x1 t2-t1 It is negative here. (x2 > x1) & (t2 > t1) Gradient = x2-x1 t2-t1 It is positive here. B (t2 x2) Velocity =0 E D displacement C M (t1, x1) B (t2, x1) A O X Velocity decreasing time Displacement Velocity constant M Velocity is constant Velocity increasing (t1, x1) displacement (t2, x2) N (t1, x2) time X Time Y N O 1- Graph OB and MN is showing that the body is moving with constant velocity. In graph OB the gradient is positive so it is positive velocity. In graph MN the gradient is negative so it is positive velocity. 2- In graph OC the displacement is decreasing with time .It shows that velocity is decreasing. 3- In graph OA the displacement is increasing with time .It shows that velocity is increasing. 4- Graph DE shows that displacement is zero so the velocity is also zero.

  29. Application of vectors With the help of vectors it is easy to comprehended the complicated chapters in higher sciences. Today vectors are used to explain higher mathematics and physics

  30. 1-Explain vectors and scalars 2- How vector method of doing questions are easy. 3- What is easy, pulling of trolley or pushing of trolley?

  31. Presented by U C Pandey Lecturer in Mathematics G.I.C. Dinapani Almora, Uttarakhand India E-mail – umeshchandrapandeya47@rediffmail.com

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