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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

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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.

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**F Italic font signifying its magnitude **

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)

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**VECTOR AND SCALAR Defining a Vector by: Cartesian Vector**

Ex. F = 59i + 59j + 29k N the magnitude is F = ( ) F = N Due to which is the vector ??

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

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**VECTOR AND SCALAR Defining a Vector by: Unit Vector F**

Ex. F = F u (use the previous example) = F for magnitude (F2 = Fx2 + Fy2 + Fz2) u for direction (dimensionless and unity) F u

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**VECTOR AND SCALAR = 0.67i + 0.67j + 0.33k**

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 88.33 u u

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**VECTOR AND SCALAR F F = 88.33 N U = 0.67i + 0.67j + 0.33k**

Z (k) F = N U U = 0.67i j k F = = = O Y (j) X (i)

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**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) r (position vector) r (position vector magnitude) u =

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**VECTOR AND SCALAR F Given: F = 150 N Required: a. F ?**

6 m 4 m 2 m Z (k) Given: F = 150 N A F Required: a. F ? b. , , ? O Y (j) X (i) U

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**VECTOR AND SCALAR Solution: r u 2i + 4j + 6k 7.48 u = = =**

F = F u r u = 2i + 4j + 6k = 7.48 u = 0.27i j +0.80k

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**VECTOR AND SCALAR Solution: a. F = F u = 150 (0.27i + 0.53j +0.80k)**

F = 40.5i j + 120k b. = cos = (angle from x-axis) = cos = (angle from y-axis) = cos = (angle from z-axis)

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**Operations of Vector Addition Subtraction Dot Product Cross Product**

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

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**VECTOR AND SCALAR Addition F2 R O F1 = Ry Rx Tan -1 R = F1 + F2 R**

(F1x + F2x) i + (F1y + F2y) j + (F1z+F2z) k

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**VECTOR AND SCALAR Addition F2 R**

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

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**VECTOR AND SCALAR Subtraction F1 O F2 R = Ry Rx Tan -1 R = F1 - F2**

(F1x - F2x) i + (F1y - F2y) j + (F1z - F2z) k

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**(the sense is opposite to the given diagram)**

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

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

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**A . B = AB cos (General Formula)**

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

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**VECTOR AND SCALAR From Example:**

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

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**VECTOR AND SCALAR The dot product is used to determine:**

The angle between the tails of the vectors. A . B = cos -1 AB The projected component of a vector V onto an axis defined by its unit vector u

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Welcome to the Jungle

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**VECTOR AND SCALAR Example: Given : Figure 1 Required: **

FBA (Magnitude) X (i) Z (k) Y (j) O B A C F = 100 N Fig.1

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**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 rBA . rBC 70000 cos = = = = 0.738 rBA rBC (300)(316.23) 94869 = Cos = o (answer)

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**VECTOR AND SCALAR Solution : FBA rBA uBA = -200i – 200j + 100k 300 =**

rBC uBC = -0i – 300j + 100k 316.2 = = – 0.949j k FBC = FBC . uBC = (– 0.949j k) = -94.9j k FBA = FBC . uBA = (-94.9i j) . (-0.667i – 0.667j k) = = 73.8 N (answer)

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**VECTOR AND SCALAR Solution : Alternative Solution**

FBA = (100 N) (cos 42.45o) = N FBA = FBA uBA = (-0.667i j k) = -49.2i – 49.2j k

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

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**VECTOR AND SCALAR A = B x C 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

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VECTOR AND SCALAR Right Hand Rule

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VECTOR AND SCALAR Right Hand Rule

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VECTOR AND SCALAR Right Hand Rule

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

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**VECTOR AND SCALAR A = B x C**

= (Bx i + By j + Bz k) x (Cx i + Cy j + Cz k) i j k Bx By Bz Cx Cy Cz i j k Bx By Bz Cx Cy Cz i j Bx By Cx Cy = = - + A = (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

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**Full caution for the +/- sign and subscripts**

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

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**VECTOR AND SCALAR Example: Given : Figure 2 Required :**

Mo (Moment at point O) My (Moment about y axis) B A F = 100N X (i) Z (k) Y (j) O Mo

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**Solution: Finding the vectors needed**

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

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**VECTOR AND SCALAR B A F = 100 N O Mo 400j 400i + 150j – 200k F**

X (i) Z (k) Y (j) O Mo 400j 400i + 150j – 200k F = 78.07i – 48.79j – 39.04k

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**VECTOR AND SCALAR Mo = OA x F i j k 0 400 0 78.07 -48.79 -39.04 = Mo**

= Mo = i – 31228k N.mm Mo = N.mm = cos-1 (-0.447) = (angle from x-axis) = cos-1 0 = (angle from y-axis) = cos = (angle from z-axis)

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**VECTOR AND SCALAR Mo = OB x F i j k 400 150 -200 78.07 -48.79 -39.04 =**

= Mo = i – 31228k N.mm Mo = N.mm = cos-1 (-0.447) = (angle from x-axis) = cos-1 0 = (angle from y-axis) = cos = (angle from z-axis)

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