1 / 40

Basic Physics

Basic Physics. Introduction. What is Physics? Give a few relations between physics and daily living experience Review of measurement and units SI, METRIC, ENGLISH. VECTOR AND SCALAR. Scalar is a quantity which only signifies its magnitude without its direction. (+ / - )

imogene-weber
Download Presentation

Basic Physics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Basic Physics

  2. Introduction What is Physics? Give a few relations between physics and daily living experience 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 FBold font F Italic font signifying its magnitude FNormal Font with an arrow head on top of it (Use this)

  5. VECTOR AND SCALAR Defining a Vector by: Cartesian Vector Ex. F = 59i + 59j + 29k N the magnitude is F = (592 + 592 + 292) F = 88.33 N Due to which is the vector ??

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

  7. u F F VECTOR AND SCALAR Defining a Vector by: Unit Vector Ex. F = F u (use the previous example) = F for magnitude (F2 = Fx2 + Fy2 + Fz2) u for direction (dimensionless and unity)

  8. u u 59i + 59j + 29k 88.33 VECTOR AND SCALAR Magnitude F = (592 + 592 + 292) F = 88.33 N Direction = = 0.67i + 0.67j + 0.33k = cos-1 0.67 = 47.9 0 (angle from x-axis) = cos-1 0.67 = 47.9 0 (angle from y-axis)  = cos-1 0.33 = 70.7 0 (angle from z-axis)

  9. U U = 0.67i + 0.67j + 0.33k VECTOR AND SCALAR Z (k) F = 88.33 N F  = 47.9 0  = 47.9 0  = 70.7 0   O  Y (j) X (i)

  10. u r(position vector) r (position vector magnitude) VECTOR AND SCALAR Defining a Vector by: 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) =

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

  12. u u F = F u r r VECTOR AND SCALAR = Solution: 2i + 4j + 6k = 7.48 = 0.27i + 0.53j +0.80k

  13. VECTOR AND SCALAR a. F=Fu = 150 (0.27i + 0.53j +0.80k) F = 40.5i + 79.5j + 120k Solution: b.  = cos-1 0.27 = 74.3 0 (angle from x-axis)  = cos-1 0.53 = 58.0 0 (angle from y-axis)  = cos-1 0.80 = 36.9 0 (angle from z-axis)

  14. VECTOR AND SCALAR Operations of Vector Addition Subtraction Dot Product Cross Product

  15. F2 F2 R R R F1 F1 = (F1x + F2x) i + (F1y + F2y) j + (F1z+F2z) k Ry  = Tan -1 Rx VECTOR AND SCALAR Addition  O = +

  16. F2 F2 R R R F1 F1 = (F1x + F2x) i + (F1y + F2y) j + (F1z+F2z) k Ry  = Tan -1 Rx VECTOR AND SCALAR Resultant is directed from initial tail towards final arrow head Addition  O = +

  17. F2 F2 R R R F1 F1 = (F1x - F2x) i + (F1y - F2y) j + (F1z - F2z) k Ry  = Tan -1 Rx VECTOR AND SCALAR O  Subtraction = -

  18. F2 F2 R R R F1 F1 = (F1x - F2x) i + (F1y - F2y) j + (F1z - F2z) k Ry  = Tan -1 Rx VECTOR AND SCALAR O  Take note and watch out !!! (the sense is opposite to the given diagram) Subtraction = -

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

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

  21. VECTOR AND SCALAR F . d =Fd cos  (Using Vectors’ magnitude) = (Fxi + Fyj + Fzk) . (dxi + dyj + dzk) = Fx dx + Fy dy + Fz dz(Using Component Vector) From Example: The dot product of two vectors is called scalar product since the result is a scalar and not a vector

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

  23. Welcome to the Jungle

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

  25. rBA. rBC VECTOR AND SCALAR Solution : • Angle  • Find position vectors from B to A and B to C • rBA = -200i – 200j + 100k • r BC = -0i – 300j + 100k = – 300j + 100k 0 + 60000 + 10000 70000 cos  = = = = 0.738 rBA rBC (300)(316.23) 94869  = Cos -1 0.738 = 42.45 o (answer)

  26. uBC = uBA = rBA rBC -200i – 200j + 100k 300 rBA rBC -0i – 300j + 100k 316.2 FBC = FBC. uBC FBA = FBC. uBA VECTOR AND SCALAR Solution : • FBA = = -0.667i – 0.667j + 0.33k = = – 0.949j + 0.316k = 100 . (– 0.949j + 0.316k) = -94.9j + 31.6k = (-94.9i + 31.6j) . (-0.667i – 0.667j + 0.33k) = 63.3 + 10.5 = 73.8 N (answer)

  27. FBA = FBA uBA VECTOR AND SCALAR Solution : Alternative Solution FBA = (100 N) (cos 42.45o) = 73.79 N = 73.79 (-0.667i - 0.667j + 0.33k) = -49.2i – 49.2j + 24.35k

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

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

  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. A = B x C i j k Bx By Bz Cx Cy Cz i j k Bx By Bz Cx Cy Cz A VECTOR AND SCALAR = (Bx i + By j + Bz k) x (Cx i + Cy j + Cz k) i j Bx By Cx Cy = = - + = (By Cz – Bz Cy)i + (Bz Cy – Bx Cz)j + (Bx Cy – By Cx)z = (By Cz – Bz Cy)i – (Bx Cz – Bz Cx)j + (Bx Cy – By Cx)z

  35. A = B x C i j k Bx By Bz Cx Cy Cz i j k Bx By Bz Cx Cy Cz A VECTOR AND SCALAR Full caution for the +/- sign and subscripts = (Bx i + By j + Bz k) x (Cx i + Cy j + Cz k) i j Bx By Cx Cy = = - + = (By Cz – Bz Cy)i + (Bz Cy – Bx Cz)j + (Bx Cy – By Cx)z = (By Cz – Bz Cy)i – (Bx Cz – Bz Cx)j + (Bx Cy – By Cx)z

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

  37. = F u F F OA OB VECTOR AND SCALAR ( ) 400i – 250j – 200k = 100 Solution: Finding the vectors needed (4002 + 2502 + 2002) = 78.07i – 48.79j – 39.04k = 400j = 400i + 150j – 200k

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

  39. Mo Mo F OA i j k 0 400 0 78.07 -48.79 -39.04 VECTOR AND SCALAR = x = = -15616i – 31228k N.mm Mo = 34914.86 N.mm  = cos-1 (-0.447) = 116.55 0 (angle from x-axis)  = cos-1 0 = 90.0 0 (angle from y-axis)  = cos-1 0.894 = 26.57 0 (angle from z-axis)

  40. Mo Mo F OB i j k 400 150 -200 78.07 -48.79 -39.04 VECTOR AND SCALAR = x = = -15616i – 31228k N.mm Mo = 34914.86 N.mm  = cos-1 (-0.447) = 116.55 0 (angle from x-axis)  = cos-1 0 = 90.0 0 (angle from y-axis)  = cos-1 0.894 = 26.57 0 (angle from z-axis)

More Related