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, 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

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:

1. 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 ??

VECTOR AND SCALAR

X (i)

Z(k)

Y(j)

F = 59i + 59j + 29k N

O

VECTOR AND SCALARDefining a Vector by:2. Unit Vector

Ex. F = F u (use the previous example)

=

F for magnitude (F2 = Fx2 + Fy

2 + Fz2)

u for direction (dimensionless and unity)

uFF

VECTOR AND SCALARMagnitude 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)

59i + 59j + 29k88.33

u

u

VECTOR AND SCALAR

X (i)

Z (k)

Y (j)

F = 88.33 N

U

FU = 0.67i + 0.67j + 0.33k

= 47.9 0

= 47.9 0

= 70.7 0

O

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 =

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

VECTOR AND SCALAR

Solution:

rr

u =

F = F u

=2i + 4j + 6k

7.48

= 0.27i + 0.53j +0.80ku

VECTOR AND SCALAR

Solution:

F = F u

= 150 (0.27i + 0.53j +0.80k)

F = 40.5i + 79.5j + 120k

= 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)

a.

b.

VECTOR AND SCALAR

Operations of Vector

1. Addition

2. Subtraction

3. Dot Product

4. Cross Product

VECTOR AND SCALAR

1. Addition

F1

RF2

R = F1 F2+

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

=Ry

Rx

Tan -1

O

VECTOR AND SCALAR

1. Addition

F1

RF2

R = F1 F2+

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

=Ry

Rx

Tan -1

O

Resultant is directed from initial tail towards final arrow head

VECTOR AND SCALAR

2. Subtraction F1

RF2

R = F1 F2-

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

=Ry

Rx

Tan -1

O

VECTOR AND SCALAR

2. Subtraction F1

RF2

R = F1 F2-

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

=

Ry

Rx

Tan -1O

Take note and watch out !!!

(the sense is opposite to the given

diagram)

VECTOR AND SCALAR

3. Dot Product

F

d

X (i)

Z (k)

Y (j)

F

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

i . j = 0 i . k = 0 k . j = 0

j . j = 1 k . k = 1

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

The dot product is used to determine:

1. The angle between the tails of the vectors.

VECTOR AND SCALAR

= cos -1A . B

AB

2. The projected component of a vector V onto an axis defined by its unit vector u

Welcome to

the

Jungle

VECTOR AND SCALAR

X (i)

Z (k)

Y (j)O

B

A

C

F = 100 N

Given : Figure 1

Required:

1.

2. FBA (Magnitude)

Fig.1

Example:

VECTOR AND SCALAR

Solution :

1. Angle

Find position vectors from B to A and B to C

rBA = -200i – 200j + 100k

r BC = -0i – 300j + 100k = – 300j + 100k

cos =rBA . rBC

rBA rBC=

0 + 60000 + 10000

(300)(316.23)=

70000

94869= 0.738

= Cos -1 0.738 = 42.45 o (answer)

VECTOR AND SCALAR

Solution :

=rBA uBA =

rBA

2. FBA

-200i – 200j + 100k

300= -0.667i – 0.667j + 0.33k

rBC uBC =

rBC =

-0i – 300j + 100k

316.2= – 0.949j + 0.316k

FBC = FBC . uBC = 100 . (– 0.949j + 0.316k) = -94.9j + 31.6k

FBA = FBC . uBA = (-94.9i + 31.6j) . (-0.667i – 0.667j + 0.33k)

= 63.3 + 10.5 = 73.8 N (answer)

VECTOR AND SCALAR

Solution :

Alternative Solution

FBA = (100 N) (cos 42.45o)

= 73.79 N

FBA = FBA uBA = 73.79 (-0.667i - 0.667j + 0.33k)

= -49.2i – 49.2j + 24.35k

VECTOR AND SCALAR

4. 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

j k

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+

-

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

=

= (By Cz – Bz Cy)i + (Bz Cy – Bx Cz)j + (Bx Cy – By Cx)zA

i j k

Bx By Bz

Cx Cy Cz

=i j

Bx By

Cx Cy

+-

= (By Cz – Bz Cy)i – (Bx Cz – Bz Cx)j + (Bx Cy – By Cx)z

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

=

= (By Cz – Bz Cy)i + (Bz Cy – Bx Cz)j + (Bx Cy – By Cx)zA

i j k

Bx By Bz

Cx Cy Cz

=i j

Bx By

Cx Cy

+-

= (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

Example: Given : Figure 2

Required :

1. Mo (Moment at point O)

2. My (Moment about y axis)

B

A

F = 100N

X (i)

Z (k)

Y (j)O

Mo

VECTOR AND SCALAR

Solution:Finding the vectors neededF = F u

= 100400i – 250j – 200k

(4002 + 2502 + 2002)( )

F = 78.07i – 48.79j – 39.04k

OA = 400j

OB = 400i + 150j – 200k

VECTOR AND SCALAR

B

A

F = 100 N

X (i)

Z (k)

Y (j)

O

Mo

400j

400i + 150j – 200k

F = 78.07i – 48.79j – 39.04k

VECTOR AND SCALAR

Mo =

=

Mo = -15616i – 31228k N.mm

FOA x

i j k

0 400 0

78.07 -48.79 -39.04

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)

VECTOR AND SCALAR

Mo =

Mo = -15616i – 31228k N.mm

FOB x

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)

i j k

400 150 -200

78.07 -48.79 -39.04

=