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Basic Physics. Introduction 1.What is Physics? 2.Give a few relations between physics and daily living experience 3.Review of measurement and units SI,

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Presentation on theme: "Basic Physics. Introduction 1.What is Physics? 2.Give a few relations between physics and daily living experience 3.Review of measurement and units SI,"— Presentation transcript:

1 Basic Physics

2 Introduction 1.What is Physics? 2.Give a few relations between physics and daily living experience 3.Review of measurement and units SI, METRIC, ENGLISH

3 VECTOR AND SCALAR Scalar is a quantity which only signifies its magnitude without its direction. (+ / - ) Ex. 1kg of apple, 273 degrees centigrade, etc. Vector is a quantity with magnitude and direction. (+ / - ) Ex. Velocity of a moving object – a car with a velocity of 100 km/hr due to North West, etc.

4 VECTOR AND SCALAR Writing conformity F Bold font F Italic font signifying its magnitude F Normal Font with an arrow head on top of it (Use this)

5 VECTOR AND SCALAR Defining a Vector by: 1.Cartesian Vector Ex. F = 59i + 59j + 29k N the magnitude is F =  ( ) F = N Due to which is the vector ??

6 VECTOR AND SCALAR X (i) Z(k) Y(j) F = 59i + 59j + 29kN O

7 VECTOR AND SCALAR Defining a Vector by: 2.Unit Vector Ex. F = F u (use the previous example) = F for magnitude (F 2 = F x 2 + F y 2 + F z 2 ) u for direction (dimensionless and unity) u F F

8 VECTOR AND SCALAR Magnitude F =  ( ) F = N Direction = = 0.67i j k  = cos = (angle from x-axis)  = cos = (angle from y-axis)  = cos = (angle from z-axis) 59i + 59j + 29k u u

9 VECTOR AND SCALAR X (i) Z (k) Y (j) F = N U F U = 0.67i j k     =  =  = O

10 VECTOR AND SCALAR Defining a Vector by: 3.Position Vector Similar to unit vector, it differs on how to locate the vector’s direction which is using the point coordinate. Ex. F = F u (see next example) r (position vector) r(position vector magnitude) u =

11 VECTOR AND SCALAR U 6 m 4 m 2 m Given: F = 150 N Required: a. F ? b. , ,  ? X (i) Z (k) Y (j) F    O A

12 VECTOR AND SCALAR Solution: r r u = F = F u = 2i + 4j + 6k 7.48 =0.27i j +0.80k u

13 VECTOR AND SCALAR Solution: F = F u = 150 (0.27i j +0.80k) F = 40.5i j + 120k  = cos = (angle from x-axis)  = cos = (angle from y-axis)  = cos = (angle from z-axis) a. b.

14 VECTOR AND SCALAR Operations of Vector 1.Addition 2.Subtraction 3.Dot Product 4.Cross Product

15 VECTOR AND SCALAR 1.Addition F1F1 R F2F2 R = F1F1 F2F2 + R = (F 1 x + F 2 x) i + (F 1 y + F 2 y) j + (F 1 z+F 2 z) k   = RyRy RxRx Tan -1 O

16 VECTOR AND SCALAR 1.Addition F1F1 R F2F2 R = F1F1 F2F2 + R = (F 1 x + F 2 x) i + (F 1 y + F 2 y) j + (F 1 z+F 2 z) k   = RyRy RxRx Tan -1 O Resultant is directed from initial tail towards final arrow head

17 VECTOR AND SCALAR 2.Subtraction F1F1 R F2F2 R = F1F1 F2F2 - R = (F 1 x - F 2 x) i + (F 1 y - F 2 y) j + (F 1 z - F 2 z) k   = RyRy RxRx Tan -1 O

18 VECTOR AND SCALAR 2.Subtraction F1F1 R F2F2 R = F1F1 F2F2 - R = (F 1 x - F 2 x) i + (F 1 y - F 2 y) j + (F 1 z - F 2 z) k   = RyRy RxRx Tan -1 O Take note and watch out !!! (the sense is opposite to the given diagram)

19 VECTOR AND SCALAR 3.Dot Product F d  X (i) Z (k) Y (j) F

20 VECTOR AND SCALAR A. B = AB cos  (General Formula)  VectorMagnitude  The angle between vectors (between their tails) Cartesian Unit vector dot product i. i = 1 i. j = 0i. k = 0k. j = 0 j. j = 1k. k = 1

21 VECTOR AND SCALAR From Example: F. d = Fd cos  (Using Vectors’ magnitude) = (F x i + F y j + F z k). (d x i + d y j + d z k) = F x d x + F y d y + F z d z (Using Component Vector) The dot product of two vectors is called scalar product since the result is a scalar and not a vector

22 The dot product is used to determine: 1.The angle between the tails of the vectors. VECTOR AND SCALAR  = cos -1 A. B AB 2. The projected component of a vector V onto an axis defined by its unit vector u

23 Welcome to the Jungle

24 VECTOR AND SCALAR X (i) Z (k) Y (j) O B A C F = 100 N  Given : Figure 1 Required: 1.  2.F BA (Magnitude) Fig.1 Example:

25 VECTOR AND SCALAR Solution : 1.Angle  Find position vectors from B to A and B to C r BA = -200i – 200j + 100k r BC = -0i – 300j + 100k= – 300j + 100k cos  = r BA. r BC r BA r BC = (300)(316.23) = =0.738  = Cos = o (answer)

26 VECTOR AND SCALAR Solution : = r BA u BA = r BA 2.F BA -200i – 200j + 100k 300 = i – 0.667j k r BC u BC = r BC = -0i – 300j + 100k = – 0.949j k F BC = F BC. u BC = 100. (– 0.949j k)= -94.9j k F BA = F BC. u BA = (-94.9i j).(-0.667i – 0.667j k) = = 73.8 N(answer)

27 VECTOR AND SCALAR Solution : Alternative Solution F BA = (100 N) (cos o ) = N F BA = F BA u BA = (-0.667i j k) = -49.2i – 49.2j k

28 VECTOR AND SCALAR 4.Cross Product B A F X (i) Z (k) Y (j) O

29 VECTOR AND SCALAR A = B x CA is equal to B cross C Apply the right hand rule i jk i x j = k j x k = i k x i = j j x i = -k k x j = -i i x k = -j i x i = 0 j x j = 0 k x k = 0 + -

30 VECTOR AND SCALAR Right Hand Rule

31 VECTOR AND SCALAR Right Hand Rule

32 VECTOR AND SCALAR Right Hand Rule

33 VECTOR AND SCALAR Right Hand Rule ……. (answer for yourself)

34 VECTOR AND SCALAR A = B x C = (B x i + B y j + B z k) x (C x i + C y j + C z k) ijkBxByBzCxCyCzijkBxByBzCxCyCz = = (B y C z – B z C y )i + (B z C y – B x C z )j + (B x C y – B y C x )zA ijkBxByBzCxCyCzijkBxByBzCxCyCz = ijBxByCxCyijBxByCxCy +- = (B y C z – B z C y )i – (B x C z – B z C x )j + (B x C y – B y C x )z

35 VECTOR AND SCALAR A = B x C = (B x i + B y j + B z k) x (C x i + C y j + C z k) ijkBxByBzCxCyCzijkBxByBzCxCyCz = = (B y C z – B z C y )i + (B z C y – B x C z )j + (B x C y – B y C x )zA ijkBxByBzCxCyCzijkBxByBzCxCyCz = ijBxByCxCyijBxByCxCy +- = (B y C z – B z C y )i – (B x C z – B z C x )j + (B x C y – B y C x )z Full caution for the +/- sign and subscripts

36 VECTOR AND SCALAR Example: Given : Figure 2 Required : 1.M o (Moment at point O) 2.M y (Moment about y axis) B A F = 100N X (i) Z (k) Y (j) O MoMo

37 VECTOR AND SCALAR Solution: Finding the vectors needed F= F u = i – 250j – 200k ( ) () F= 78.07i – 48.79j – 39.04k OA= 400j OB= 400i + 150j – 200k

38 VECTOR AND SCALAR B A F = 100 N X (i) Z (k) Y (j) O MoMo 400j 400i + 150j – 200k F= 78.07i – 48.79j – 39.04k

39 VECTOR AND SCALAR MoMo = = MoMo = i – 31228kN.mm FOA x i jk MoMo = N.mm  = cos -1 (-0.447) = (angle from x-axis)  = cos -1 0 = (angle from y-axis)  = cos = (angle from z-axis)

40 VECTOR AND SCALAR MoMo = MoMo = i – 31228kN.mm FOB x MoMo = N.mm  = cos -1 (-0.447) = (angle from x-axis)  = cos -1 0 = (angle from y-axis)  = cos = (angle from z-axis) i jk =


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