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##### Basic Physics

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**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. (+ / - ) 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.**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)**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 ??**F = 59i + 59j + 29k N**VECTOR AND SCALAR Z(k) O Y(j) X (i)**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)**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)**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)**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) =**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)**u**u F = F u r r VECTOR AND SCALAR = Solution: 2i + 4j + 6k = 7.48 = 0.27i + 0.53j +0.80k**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)**VECTOR AND SCALAR**Operations of Vector Addition Subtraction Dot Product Cross Product**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 = +**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 = +**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 = -**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 = -**F**F Z (k) X (i) d Y (j) VECTOR AND SCALAR Dot Product**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**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**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.**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**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)**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)**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**Z (k)**O A Y (j) X (i) B F VECTOR AND SCALAR Cross Product**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**VECTOR AND SCALAR**Right Hand Rule**VECTOR AND SCALAR**Right Hand Rule**VECTOR AND SCALAR**Right Hand Rule**VECTOR AND SCALAR**Right Hand Rule ……. (answer for yourself)**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**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**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:**= 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**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**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)**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)