Fisika Dasar - Introduction and Vector 1.ppt

8/11/2019 Fisika Dasar - Introduction and Vector 1.ppt

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Basic Physics


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Introduction

1. What is Physics?

2. Give a few relations between physicsand daily living experience

3. Review of measurement and units SI,METRIC, ENGLISH


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VECTOR AND SCALAR

Scalar is a quantity which only signifies itsmagnitude 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 carwith a velocity of 100 km/hr due to NorthWest, etc.


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VECTOR AND SCALAR

Writing conformity

F Bold fontF Italic font signifying its magnitude F Normal Font with an arrow

head on top of it (Use t h is )


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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 ??


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VECTOR AND SCALAR

X (i)

Z(k)

Y(j)

F = 59i + 59j + 29k N

O


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VECTOR AND SCALARDefining a Vector by:2. Unit Vector

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

=

F for magnitude (F 2 = F x2 + F y2 + F z2)

u for direction (dimensionless and unity)

uF F


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VECTOR AND SCALARMagnitude F = (59 2 + 59 2 + 29 2)

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 + 29k88.33u

u


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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


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VECTOR AND SCALAR

Defining a Vector by:3. Position Vector

Similar to unit vector, it differs on how tolocate the vectors direction which isusing 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

U

6 m

Given:

F = 150 N

Required:

a. F ?

b. , , ?

X (i )

Z (k )

Y ( j )

F

O

A


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VECTOR AND SCALAR

Solution:

r r

u =

F = F u

=2i + 4j + 6k

7.48 = 0.27i + 0.53j +0.80ku


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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.


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VECTOR AND SCALAR

Operations of Vector1. Addition

2. Subtraction3. Dot Product4. Cross Product


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VECTOR AND SCALAR

1. Addition

F1

RF2

R = F1 F2+

R = (F 1x + F 2x) i + (F 1y + F 2y) j + (F 1z+F 2z) k

=R y

RxTan -1

O


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VECTOR AND SCALAR

1. Addition

F1

RF2

R = F1 F2+

R = (F 1x + F 2x) i + (F 1y + F 2y) j + (F 1z+F 2z) k

=R y

RxTan -1

O

Resultant is directedfrom initial tail towardsfinal arrow head


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VECTOR AND SCALAR

2. SubtractionF1

RF2

R = F1 F2-

R = (F 1x - F 2x) i + (F 1y - F 2y) j + (F 1z - F 2z) k

= RyRx

Tan -1

O


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VECTOR AND SCALAR

2. SubtractionF1

RF2

R = F1 F2-

R = (F 1x - F 2x) i + (F 1y - F 2y) j + (F 1z - F 2z) k

=

Ry

RxTan -1O

Take note and watchout !!!

(the sense isopposite to the given

diagram)


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VECTOR AND SCALAR

3. Dot Product

F

X (i )

Z (k )

Y ( j )

F


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VECTOR AND SCALAR

A . B = AB cos (General Formula)

Vector Magni tude

The ang le betw eenvectors (be tweenth eir tai ls )

Cartesian Unit vector dot product

i . i = 1

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

j . j = 1 k . k = 1


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VECTOR AND SCALAR

From Example: F . d = Fd cos (Using Vectors magnitude)

= (F xi + F y j + F zk) . (d xi + d y j + d zk)

= F x dx + F y d y + F z d z (Using Component Vector)

The dot product of two vectors is called scalar productsince the result is a scalar and not a vec tor


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The dot produ ct is used to dete rmine:1. The angle between the tails o f the vectors .

VECTOR AND SCALAR

= cos-1

A . B

AB2. The projecte d component of a vector V onto an axis

defined by it s unit vector u


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VECTOR AND SCALAR

X (i )

Z (k )

Y ( j )O

B

A

C

F = 100N

Given : Figure 1

Required:

1.

2. F BA (Magnitude)

Fig.1

Example:


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VECTOR AND SCALAR

Solution :

1. Angle

Find position vectors from B to A and B to C

r BA = -200i 200j + 100k

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

cos =r BA . r BC r BA r BC

=0 + 60000 + 10000

(300)(316.23)=

7000094869

= 0.738

= Cos -1 0.738 = 42.45 o (answer)


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VECTOR AND SCALAR

Solution :

=r BAuBA =

r BA

2. F BA

-200i 200j + 100k300 = -0.667i 0.667j + 0.33k

r BCuBC =

r BC=

-0i 300j + 100k316.2

= 0.949j + 0.316k

FBC = F BC . u BC = 100 . ( 0.949j + 0.316k) = -94.9j + 31.6k

FBA = F BC . u BA = (-94.9i + 31.6j) . (-0.667i 0.667j + 0.33k)

= 63.3 + 10.5 = 73.8 N (answer)


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VECTOR AND SCALAR

Solution : Alternative Solution

FBA = (100 N) (cos 42.45o)

= 73.79 N

FBA = F BA u BA = 73.79 (-0.667i - 0.667j + 0.33k)

= -49.2i 49.2j + 24.35k


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VECTOR AND SCALAR

4. 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

j k

i x j = k

j x k = ik x i = j

j x i = -k

k x j = -ii x k = -j

i x i = 0

j x j = 0k x k = 0+

-


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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

= (B x i + B y j + B z k ) x (C x i + C y j + C z k )

i j kBx By Bz

Cx C y C z

=

= (B y C z Bz C y)i + (B z C y Bx C z) j + (B x C y By C x)z A

i j kBx By Bz

Cx C y C z

=i jBx By

Cx C y

+-

= (B y C z Bz C y)i (B x C z Bz C x) j + (B x C y By C x)z


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VECTOR AND SCALAR

A = B x C

= (B x i + B y j + B z k ) x (C x i + C y j + C z k )

i j kBx By Bz

Cx C y C z

=

= (B y C z Bz C y)i + (B z C y Bx C z) j + (B x C y By C x)z A

i j kBx By Bz

Cx C y C z

=i jBx By

Cx C y

+-

= (B y C z Bz C y)i (B x C z Bz C x) j + (B x C y By C x)z

Full cautionfor the +/- signand

subscripts


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VECTOR AND SCALAR

Example: Given : Figure 2

Required :

1. M o (Moment atpoint O)

2. M y (Moment

about y axis)

B

A

F = 100N

X (i )

Z (k )

Y ( j )O

Mo


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VECTOR AND SCALAR

Solution:Finding the vectors neededF = F u

= 100400i 250j 200k

(400 2 + 250 2 + 200 2)( )

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

X (i )

Z (k )

Y ( j )

O

Mo400i + 150j 200k

F = 78.07i 48.79j 39.04k


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VECTOR AND SCALAR

Mo =

=

Mo = -15616i 31228k N.mm

F OA 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)


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VECTOR AND SCALAR

Mo =

Mo = -15616i 31228k N.mm

F OB 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

=

Fisika Dasar - Introduction and Vector 1.ppt

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