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