chapter 2_ scalar and vector.pdf

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    Scalar and Vector

    1

    CHAPTER 2

    PHY 130

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    LESSON OUTCOMESAfter completing this chapter, you should be able to

    O state the definition of scalar and vector quantities and

    give examples of each.

    O state the differences between scalar & vector quantities.O determine the components of a given vectors

    O find the resultant of two or more vectors.

    O calculate vector addition & subtraction by using

    geometrical method & unit vector method / resolvingmethod.

    2

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    OScalarquantity : quantity with magnitude only.

    O e.g. mass, time, temperature, pressure, electric

    current, work, energy and etc.

    O Mathematics operational : ordinary algebra

    O Vector quantity:quantity with both magnitude

    direction.

    O e.g. displacement, velocity, acceleration, force,

    momentum, electric field, magnetic field and etc.

    O Mathematics operational : vector algebra

    3

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    VE TORS

    s

    4

    O Written form (notation) of vectors.

    O Notation of magnitude of vectors.

    Vector ALengthof an arrowmagnitudeof vector A

    displacement velocity acceleration

    v

    a

    s av

    aa

    s (bold) v (bold) a (bold)

    Directionof arrow directionof vector A

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    P

    5

    O Two vectorsequal if both magnitude and directionare the same.

    O If vector A is multiplied by a scalar quantity kO Then, vector A is

    O if k= +ve, the vector is in the same directionas vector A.

    O if k = - ve, the vector is in the opposite directionof vector A.

    Q

    QP

    Ak

    Ak

    A

    A

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    6

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    Addition of Vectors

    O There are two methods involved in addition of vectors graphically i.e.

    OParallelogram

    O Triangle

    O For example :

    7

    Parallelogram Triangle

    B

    A

    B

    A

    BA

    O

    BA

    B

    A

    BA

    O

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    Subtraction of Vectors

    O For example :

    8

    Parallelogram Triangle

    DC

    O

    DC

    O

    D

    DCDC

    C

    D

    DC

    C

    D

    DC

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    Resolving a Vector

    9

    yD

    xD

    0x

    y

    D

    Dx cos DDx

    cos

    D

    Dysin DDy sin

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    10

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    11

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    12

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    O The magnitude of vector D:

    O Direction of vectorD:

    O VectorDin terms of unit vectors written as

    13

    or

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    14

    A car moves at a velocity of 50 m s-1in a direction north 30

    east. Calculate the component of the velocity

    a) due north. b) due east.

    Solution :

    Example :

    N

    EW

    S

    Nv

    Ev

    v30

    60

    a)

    b)

    30vvN cos

    1sm43.3 Nv

    3050vN cos

    or60vvN sin

    6050vN sin

    30vvE sin

    1sm25 Ev

    3050vE sinor

    60vvE cos

    6050vE cos

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    15

    A particle S experienced a force of 100 N as shown in figure above. Determine

    the x-component and the y-component of the force.

    Solution :

    Example :

    120

    F

    Sx

    12060

    F

    Sx

    y

    y

    F

    xF

    Vector x-component y-component

    60FFx cos

    N50xF

    60100Fx cos

    orF

    120FFx cos

    N50x

    F

    120100Fx cos

    60FFy sin

    N86.6yF

    60100Fy sin

    or120FFy sin

    N86.6

    yF

    120100Fy sin

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    16

    The figure above shows three forces F1, F2and F3 acted on a particle O.

    Calculate the magnitude and direction of the resultant force on particle O.

    Example : y

    45o

    O

    )( N30F2

    )( N10F1

    30o x

    )( N40F3

    20

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    17

    Solution :

    O

    y

    x

    3F

    45

    o

    30o

    20

    1F

    y1F

    2F

    y2F

    x1F

    y3F

    x3F

    321r FFFFF

    yxr FFF

    x3x2x1x FFFF

    y3y2y1y FFFF

    x2F

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    18

    Solution :

    Vector x-component y-component

    20FF 1x1 cos

    1F

    3F

    2F

    2010Fx1 cosN9.40x1F

    20FF 1y1 sin2010Fy1 sin

    N3.42y1F4530Fx2 cos

    N21.2x2F

    4530Fy2 sinN21.2y2F

    3040Fx3 cos

    N34.6x3F

    3040Fy3 sin

    N20.0

    y3F

    Vector

    sum

    34.621.29.40 xFN4.00 xF

    20.021.23.42 yFN37.8 yF

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    19

    y

    xO

    Solution :

    The magnitude of the resultant force is

    and its direction is

    22 yxr FFF

    N38.0rF

    22 37.84.00 rF

    x

    y

    F

    F

    1tan

    iseanticlockwaxis-xpositivefrom264or84.0

    4.00

    37.8tan

    1

    rF

    yF

    xF

    84.0

    264

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    Unit Vectors

    O notations

    O has a magnitude of one with no units,

    O for any positional vector A, its unit vector is given by

    O Unit vector for 3 dimension axes :

    20

    A

    a

    cba ,,

    A

    a

    A

    )(@- boldjjaxisy

    )(@- boldiiaxisx

    )(@- boldkkaxisz

    1a

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    ab

    m542 kjia

    21

    Two vectors are given as:

    Calculate

    a) the vector and its magnitude,

    b) the vector and its magnitude,

    c) the vector and its magnitude.

    Solution :

    a)

    The magnitude,

    Example :

    ba

    m87 kjib

    ibaba xxx572

    jbaba yyy

    484

    m645 kjiba

    kbaba zzz 615

    m8.78645222

    ba

    ba

    2

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    22

    b)

    The magnitude,

    c)

    The magnitude,

    iabab xxx927

    jabab yyy

    1248

    m4129 kjiab

    kabab zzz 451

    m15.54129 222 ab

    ibaba xxx 372222

    jbaba yyy 084222

    m1132 kiba

    kbaba zzz 1115222

    m11.41132

    22 ba