Lec 1 - Fundamental Concepts, Force Vectors

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  • ES 11 Lecture 1

  • the study of the relationship among

    forces and their effects on bodies.

    the science which describes and predicts the conditions for rest and motion of bodies under the action of forces.

  • Mechanics

    Rigid Bodies

    Statics

    Dynamics

    Deformable Bodies

    Fluids

    Compressible

    Incompressible

  • represents the action of one body on another

    may be exerted by actual contact or at a distance

    characterized by its: Point of application

    Magnitude

    Line of action

    represented by a vector.

  • Development of other forces Reactions

    Internal forces

    Deformation of the body

    Acceleration of the body

    Applied Force

  • Applied Force

    Reaction

    Development of force or forces at

    points of contact with other

    bodies (reactions).

  • Development of forces within the

    body itself (internal forces) A A

    Applied Force

  • Applied Force

    Deformation of the body

  • Applied Force

    Acceleration of the body

  • 10 N

    30o

    10 N 30o

    Point of application (forces acting on the same particle have the same point of application)

    Magnitude Direction Line of action (angle w.r.t. a fixed axis) Sense

  • Treatment of bodies as particles - the shape and size of the object does not significantly affect the solution of the problems under consideration.

    Rigid Bodies - the problems considered in this course are assumed to be non-deformable.

  • Acceleration due to gravity

    g = 9.81 m/s^2 or g = 32.2 ft/s^2

    Quantity SI English

    Length m (meter) ft (feet)

    Mass kg (kilogram) slugs

    Time s (seconds) s (seconds)

    Force kg m/s2 OR N (newtons)

    lbs (pounds)

  • P

    Q

    A

    R

    A

    Single equivalent force having the same effect as the original forces acting on the particle

  • Parallelogram Law

    The resultant is the diagonal of the parallelogram with the two forces as its sides

    Triangle Law

    Derived from the parallelogram law

    If the two vectors are placed tip-to-tail, the resultant is the third side of the triangle

    P

    Q A R

    P

    A

    Q A P

    Q

  • P

    Q A P+Q = R

    Q QPR

    BPQQPR

    cos2222

    Law of cosines,

    Law of sines,

    P

    C

    R

    B

    Q

    A sinsinsin

    Vectors are defined as mathematical expressions possessing magnitude and direction, which add according to the parallelogram law.

    Forces are sliding vectors

  • Is vector addition commutative?

    Is vector addition associative?

  • Is vector addition commutative? The addition of vector is commutative.

    P

    Q A

    P+Q

    Q

    PQQP

    P

    Q A

    Q+P

    P

  • Adding more than 2 vectors If the vectors are coplanar, the resultant may be obtained by using the polygon rule for the addition of vectors arrange the given vectors in a tip-to-tail fashion and connect the tail of the first vector with the tip of the last one

    P

    Q A

    Q

    S

    R

    P

    Q A

    Q

    S

    R

    P+Q

  • Is vector addition associative? Vector addition is associative

    P

    A

    Q

    S

    R

    P+Q

    P

    A

    Q

    S

    R

    Q+S

    (P+Q)+S

    P+ (Q+S)

    P+Q+S = (P+Q)+S = P+ (Q+S)

  • PPP 2

    Pn

    - have the same direction as P with magnitude Pn

    P 1.5P -2P

    P

    P

    - Product nP of a positive integer n and a vector P

  • Concurrent forces: set of forces which all pass through the same point. A set of concurrent forces applied to a particle may be replaced by a single resultant force which is the vector sum of the applied forces.

  • The two forces act on a bolt at

    A. Determine their resultant.

    SOLUTION:

    Trigonometric solution - use the triangle rule for vector addition in conjunction with the law of cosines and law of sines to find the resultant.

  • Trigonometric solution - Apply the triangle rule.

    From the Law of Cosines,

    155cosN60N402N60N40

    cos222

    222 BPQQPR

    N73.97R

  • From the Law of Sines,

    A

    A

    R

    QBA

    R

    B

    Q

    A

    20

    04.15N73.97

    N60155sin

    sinsin

    sinsin

    04.35

  • 25

    Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle if the resultant of the two forces applied to the hook support is to be horizontal, (b) the corresponding magnitude of R.

    50 N

    25O

    P

  • 26

    50 N

    25O

    P

    R

    R

    O sin

    35

    25sin

    50

    sin

    (a) Determining &

    OO

    14.3735

    25sin50sin 1

    OO 25180

    O86.117

  • NR 22.73

    27

    50 N

    25O

    P

    R

    R

    O sin

    35

    25sin

    50

    sin

    (b) Determining R

    sin

    sin50R

  • Two forces are applied as shown to a hook support. Knowing that the magnitude of the resultant of the two forces is a 50 N horizontal force, determine the value of for which the applied force 2 is minimum.

    50 N

    25O

    P

    Force 1

    Force 2

  • It has been shown that the resultant of forces acting at the same point (concurrent forces) can be found.

    In the same way, a given force F can be resolved into components.

    29

  • Vector force components: two or more force vectors which, together, have the same effect as a single force vector.

  • How many components could a force be resolved into?

    Infinite number of possible components

    For a force resolved into two components,

  • 32

    15 . 0

    40 . 0

    45 . 0

    a

    b

    P Given P is 800 N, determine the components of the force in a and b axes.

  • 4045180

    33

    15 . 0

    40 . 0

    45 . 0

    a

    b

    P 15.0O 25.0O P=800N

    95O

    40O

    45O

    AF

    BF

    A

    O

    B

    OO

    FF

    40sin45sin

    800

    95sin

    NFB 85.567

    NFA 19.516

  • A force vector may be resolved into perpendicular components

  • j F F y y

    j

    i i F F x x

    F

    q

    y

    x

    i j and - unit vectors of magnitude 1 directed along the

    positive x and y axes, respectively.

    xF yF - vector components of

    FFx , Fy scalar components of

    F

    - may be positive or negative

    depending upon the sense of xF

    and yF

    - the absolute values are equal

    to the magnitude of the

    component forces and xF yF

    qcosFFx q sinFFy

    F = Fxi +Fy j

  • 0 35

    0 145 q

    N F 800

    x F

    y F FX = -655.3 N i

    FY = 458.9 N j

  • 2 - 37

    SQPR

    Wish to find the resultant of 3 or more concurrent forces,

    jSQPiSQPjSiSjQiQjPiPjRiR

    yyyxxx

    yxyxyxyx

    Resolve each force into rectangular components

    x

    xxxx

    F

    SQPR

    The scalar components of the resultant are equal to the sum of the corresponding scalar components of the given forces.

    y

    yyyy

    F

    SQPR

    x

    yyx

    R

    RRRR 122 tan q

    To find the resultant magnitude and direction,

  • 38

    tan q = x

    y

    F

    F

    F = 22

    yx FF q

    F

    x F

    y F

    x

    y

    qR

    R

    x R

    y R

    x

    y

    xx FR

    yy FR xy

    RR

    Rqtan

  • Determine the resultant of the three forces below

    39

    25o

    45o 350 N

    800 N

    600 N

    60o

    y

    x

  • 40

    25o

    45o 350 N

    800 N

    600 N

    60o

    y

    x

    RX = F x = 350 cos 25O + 800 cos 70O -600 cos 60O

    RX = 317.2 + 273.6 300 = 290.8 N

    RY = F y = 350 sin 25 + 800 sin 70O +600 sin 60O

    RY = 147.9 + 751 + 519.6 = 1419.3 N

    F= 290.8 N i +1419.3 N j

    Resultant, F

    NF 14493.14198.290 22 O4.788.290

    3.1419tan 1 q

    F = 1449 N 78.4O

  • The resultant of concurrent forces acting on a particle in space will also act at the same particle.

    Only the magnitude and direction are to be determined.

    y

    z

    x

    o

    F1

    F2

    F3

    Note: In this illustration, the magnitude and direction of all of the forces are given.

  • From the force polygon (warped), the resultant can be drawn from the tail of the first force to the head of the last force.

    The magnitude and direction of the resultant can be computed using successive use of the triangle law.

    y

    z

    x

    o

    F1

    F2 F3

    z q

    x q

    y q

    R

    Note: The sine and cosine laws are hard to implement because usually the given angles are absolute.

  • The rectangular components of a force can be determined easily depending on the given characteristics of the force.

    Given the Magnitude and Two Angles

    y

    z

    x

    o

    F

    zq

    yq

    y

    z

    x

    o

    F

    zq

    xyq

    xyF

    zF

    zxy FF qsin

    zz FF qcos

  • xyzxyxyy FFF qqq cossincos

    xyzxyxyx FFF qqq sinsinsin

    Given the Magnitude and Two Angles

    y

    z

    x

    o

    F

    zq

    xyq

    xyF

    zF

    zxy FF qsin

    y

    z

    x

    o

    xyq

    xyF

    zF

    xF

    yF

  • zz FF qcos

    Given the Magnitude and Two Angles

    y

    z

    x

    o

    xyq

    xyF

    zF

    xF

    yF xyzxyxyy

    xyzxyxyx

    FFF

    FFF

    qqq

    qqq

    cossincos

    sinsinsin

    222

    FzFyFxF

    kFzjFyiFxF

    In vector form,

  • Given the Magnitude and Three Absolute Angles

    Fx = Fcos qx Fy = F cos qy Fz = Fcos qz where cos qx, cos qy and cos qz are direction cosines F = Fxi + Fyj + Fzk

    y

    z

    x

    o

    x q

    F

    y q

    z q

  • A force of 500N forms angles of 600, 450 and 1200, respectively, with the x,y and z axes. Find the components Fx, Fy and Fz of the force. Find also the vector representation of the force.

    Fx = F cos qx = (500N)cos 600 Fx = +250N Fy = F cos qy = (500N) cos 450 Fy = +354N Fz = F cos qz = (500N) cos 1200 Fz = -250N

    y

    z

    x

    o

    F

    qx qy

    qz

    Fx

    Fy

    Fz

    F = 250N i + 354N j 250N k

  • y

    z

    x

    o

    F

    qx qy

    qz

    Fx

    Fy

    Fz

    F = 250N i + 354N j 250N k

    Note: The angle a force F forms with an axis should be measured from the positive side of the axis and will always be between 0 and 1800.

  • kzjyixFF coscoscos qqqlet = cos qx i + cos qy j+ cos qz k

    = unit vector x= cos qx y= cos qy z= cos qz

    x2 + y

    2 + z2 = 1

    cos2 qx + cos2 qy + cos

    2 qz = 1

    it follows that,

    F = F The force vector is equal to the product of the magnitude of the force and the unit vector.

  • A force has the components Fx = 20N, Fy = -30N, Fz = 60N. Determine its magnitude F and the angles qx, qy, qz it forms with the coordinate axes.

    y

    z

    x

    o

    Fx

    Fy

    Fz 222 FxFyFxF

    NNF

    NNNF

    704900

    )60()30()20( 222

  • qx = cos-1 (Fx / F) = cos

    -1 (20/70)

    qx = 73.40

    qy = cos-1 (Fy / F) = cos

    -1 (-30/70)

    qy = 115.40

    qz = cos-1 (Fz / F) = cos

    -1 (60/70)

    qy = 31.00

    y

    z

    x

    o

    Fx

    Fy

    Fz F

    F = 70 N

  • Given the Magnitude and Two Points where Force Passes

    222 dzdydxd

    zzd

    yyd

    xxd

    oez

    oey

    oex

    Fd

    dxFx F

    d

    dyFy F

    d

    dzFz

    kFzjFyiFxF

    y

    z

    x

    o

    e e e z y x E , ,

    o o o z y x O , ,

    F

  • kFzjFyiFxF

    y

    z

    x

    o

    eee zyxE ,,

    eee zyxO ,,

    F

    FxFy

    Fz

    d

    dxx qcos

    d

    dyy qcos

    d

    dzz qcos

  • Determine the vector representation of the given force.

    z

    x

    O

    2.4m E

    1.5m

    F=1.6kN

    1.2m

    1.5m

    y

  • EXAMPLE 1-8

    z

    x

    O

    2.4m E

    1.5m

    F=1.6kN

    1.2m

    1.5m

    y

    O(1.2, 1.5, 0.0) E(0.0, 2.4, 1.5)

    dx = 0.0 -1.2 = -1.2 dy = 2.4 -1.5 = +0.9 dz = 1.5 - 0.0 = +1.5

    222 5.19.02.1 d

    d = 2.12m

  • kNkNFx 905.0)6.1(12.2

    2.1

    kNkNFy 679.0)6.1(12.2

    9.0

    kNkNFz 131.1)6.1(12.2

    5.1

    kkNjkNikNF 131.1679.0905.0

    Fx

    Fz

    z

    x

    O

    E

    y

    Fy

  • The resultant R of two or more forces in space will

    be determined by summing their rectangular

    components. Graphical or trigonometric methods are

    generally not practical in the case of forces in space.

    FR FzFyjFxiRzkRyiRxi kFzjFyiFx

    FxRx FyRy FzRz

  • 222 RzRyRxR

    R

    Rxx qcos

    R

    Ryy qcos

    R

    Rzz qcos

  • Determine graphically, the magnitude and direction of the resultant of the two forces using (a) Parallelogram law and (b) the triangle rule.

    900 N

    600 N

    30o

    45o

  • A parallelogram with sides equal to 900 N and 600 N is drawn to scale as shown.

    From the scaled drawing of the forces, the resultant is

    R 1400 N 900 N

    600 N

    30o

    45o

    R

    q q 46o

    Note: The triangle rule may also be used. Join the forces in a tip to tail fashion and measure the magnitude and direction of the resultant.

  • For the magnitude of R, using the cosine law:

    oR 135cos6009002600900 222

    NR 13916.1390

    For angle q, using sine law:

    sin

    600

    135sin

    O

    R

    900 N

    600 N

    30o

    135o

    R

    q

    OO 8.1730 qThe angle of the resultant:

    OO

    8.171391

    135sin600sin 1

  • Sample Problem 2.10 : (2.89) A frame ABC is supported in part by cable DBE which passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support D.

    400mm

    210mm 280mm

    510mm

    480mm 600mm

    x z

    A

    B

    C

    D E

  • O(0, 510, 280)

    E(480, 0, 600)

    dx = xE xO = 480-0=480

    dy = yE yO = 0-510=-510

    dz = zE zO = 600-280=320

    d = 770 mm

    400mm

    210mm

    280mm

    510mm

    480mm

    600mm

    x z

    A

    B

    C

    D E

    TDB = 385N

    TDBY

    TDBX TDBZ

    E(480, 0, 600)

    O(0, 510, 280)

    NTd

    dT DB

    x

    DBX 240385770

    480

    NTd

    dT DB

    y

    DBY 255385770

    510

    NTd

    dT DB

    zDBZ 160385

    770

    320

    NkjikTjTiTT DBZDBYDBXDB 160255240

  • Solution:

    Position vector of BH = 0.6 m i + 1.2 m j - 1.2 m k

    Magnitude, BH = 0 6 1 2 1 2 1 8 2 2 2 . . . . m

    BH

    BH BH BH BH

    BH

    x y z

    BH

    BH m i m j m k

    T T T BH

    BH

    N

    m m i m j m k

    T N i N j N k

    F N F N F N

    | | . ( . . . )

    | | . | | | | .

    . . .

    ( ) (500 ) (500 )

    1

    1 8 0 6 1 2 1 2

    750

    1 8 0 6 1 2 1 2

    250

    250 500 500

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