# Basic Physics.

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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. (+ / - ) 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.

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)

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 ??

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

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

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

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)

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 =

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

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

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)

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

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

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

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

(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

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

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

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

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

Welcome to the Jungle

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

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)

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)

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

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

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

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)

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

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

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

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

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

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)

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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