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Applied Mechanics 12104

Unit 1. Vectors and Scalars

Facilitator: Ms. Maloisane

Contents
  • Mechanics

  • Fundamental Concepts

  • Units of Measurement

  • The International System of Units

  • Scalars and Vectors

  • Vector Operations

  • Vector Addition of Forces

What is Mechanics?
  • Definition: Mechanics is the science which describes and predicts the conditions of rest or motion of bodies under the action of forces.

Branches of Mechanics
  1. Mechanics

    • Rigid Body Mechanics: Study of bodies that do not deform under applied forces.

    • Deformable Body Mechanics: Study of bodies that deform when forces are applied.

    • Fluids Mechanics: Study of fluids (liquids and gases) and the forces acting upon them.

  2. Statics:

    • Concerned with bodies at rest or moving with constant velocity.

  3. Dynamics:

    • Study of accelerated motion of bodies.

  4. Kinematics:

    • Concerned with the geometric aspects of motion—describing movement without considering the forces involved.

  5. Kinetics:

    • Concerned with the forces causing motion.

Fundamental Concepts: Basic Quantities
  • Four fundamental physical quantities:

    • Length: Locates the position of a point in space.

    • Mass: Measure of a quantity of matter.

    • Time: Succession of events.

    • Force: A "push" or "pull" exerted by one body on another.

Fundamental Concepts: Idealizations
  • Particles: Has mass, and size can be neglected.

  • Rigid Body: A combination of a large number of particles treated as one entity.

  • Concentrated Force: The effect of a loading treated as applied at one point.

Newton's Three Laws of Motion
  1. First Law (Law of Inertia):

    • A particle originally at rest, or moving in a straight line with constant velocity, will remain in this state provided that the particle is not subjected to an unbalanced force.

    Figure illustrating equilibrium and forces
  2. Second Law:

    • A particle acted upon by an unbalanced force F experiences an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force.

    • Mathematical Representation: F=maF = ma

  3. Third Law:

    • The mutual forces of action and reaction between two particles are equal, opposite, and collinear.

    • Action-reaction example: Force of A on B equals the force of B on A.

Units of Measurement
  • Length: Meter (m)

  • Time: Second (s)

  • Mass: Kilogram (kg)

  • Force: Newton (N)

The International System of Units (SI)

Exponential Form

Prefix

SI Symbol

Multiple

1,000,000,000

Giga

G

10^9

1,000,000

Mega

M

10^6

1,000

Kilo

k

10^3

0.001

Milli

m

10^-3

0.000001

Micro

μ

10^-6

0.000000001

Nano

n

10^-9

Scalars and Vectors
  • Characteristics:

    • Scalar:

    • Has magnitude but not direction.

    • Example of Scalars: Mass, Volume, Length.

    • Vector:

    • A quantity that has both magnitude and direction.

    • Examples of Vectors: Force, Velocity, Acceleration, Moment.

    • Special Notation:

    • Scalars indicated by letters in italic (e.g., A).

    • Vectors represented by a letter with an arrow over it or boldface letter (e.g., $ extbf{A}$).

    • Magnitude of vector denoted as |A| or simply A.

Vector Operations
  • Multiplication and Division of a Vector:

    • Equal vectors have the same magnitude and direction.

    • A negative vector of a given vector has the same magnitude but the opposite direction.

Vector Addition
  1. Parallelogram Law:

    • Addition of two vectors A and B gives a resultant vector R.

    • The resultant is represented by the diagonal of the formed parallelogram.

  2. Triangle Construction:

    • Result R can be found by triangle construction.

    • Example: R = A + B = B + A.

  3. Special Case:

    • Vectors A and B are collinear (both have the same line of action).

Vector Subtraction
  • Considered a special case of addition:

    • Example: R=AB=A+(B)R' = A - B = A + (-B)

    • Rules of Vector Addition apply.

Vector Addition of Forces
  • Finding a Resultant Force:

    • The parallelogram law is used to find the resultant force when multiple forces act.

Procedure for Analysis (Parallelogram Law)
  1. Make a sketch using the parallelogram law.

  2. The two component forces add to form the resultant force.

  3. Resultant force is shown by the diagonal of the parallelogram.

  4. The components are shown as the sides of the parallelogram.

  5. To resolve a force into components along two axes, directed from the tail of the force:

    • Start at the head, constructing lines parallel to the axes.

    • Label all known and unknown force magnitudes and angles.

    • Identify the unknown components.

Procedure for Analysis (Trigonometry)
  1. Redraw half of the parallelogram.

  2. Magnitude of the resultant force can be determined using the law of cosines.

  3. Direction of the resultant force can be determined using the law of sines.

Adding Multiple Forces
  • When two or more forces are combined, successive applications of the parallelogram law are used to find the resultant.

    • Example: For forces F1, F2, and F3 acting at a point O:

    • First, find resultant of F1 + F2, and then add F3 to obtain the final resultant: FR=(F1+F2)+F3F_R = (F1 + F2) + F3.

Example 1.7.1
  • Scenario: The screw eye is subjected to two forces F1 and F2. Determine graphically the magnitude and direction of the resultant force.

Solution (Using the Parallelogram Law)
  • Unknowns:

    • Magnitude of the resultant force F_R and angle θ.

  • Given:

    • Force F1 = 150 N

    • Force F2 = 100 N

  • Components Calculation:

    • $x$-component (F1): 150 N cos 10° = 147.72 N

    • $y$-component (F1): 150 N sin 10° = 26.05 N

    • $x$-component (F2): 100 N cos 15° = 96.59 N

    • $y$-component (F2): 100 N sin 15° = 25.88 N

  • Summation of Components:

    • $ ext{Σ}F_x = 147.72 N + 96.59 N = 244.31 N$

    • $ ext{Σ}F_y = 26.05 N + 25.88 N = 51.93 N$

    • Resultant Magnitude:
      F<em>R=extΣF</em>x2+extΣFy2=ext[(244.31N)2+(51.93N)2]=249.78NF<em>R = ext{Σ}F</em>x^2 + ext{Σ}F_y^2 = ext{√}[(244.31 N)² + (51.93 N)²] = 249.78 N

    • Angle θ:
      θ=an1(rac51.93244.31)=12.11°θ = an^{-1}\bigg( rac{51.93}{244.31}\bigg) = 12.11°

Solution (Using Trigonometry)
  • Use of the Law of Cosines:
    FR=ext(150N2+100N22(150N)(100N)extcos(115°))=212.55NF_R = ext{√}(150 N² + 100 N² - 2(150 N)(100 N) ext{cos}(115°)) = 212.55 N

  • Use of the Law of Sines for respective angles:
    rac150NFR=racextsin115°extsinθrac{150 N}{F_R} = rac{ ext{sin} 115°}{ ext{sin} θ}

Class Exercise
  • Task: Two forces are applied to an eye bolt fastened to a beam as shown in Figure 1. With the aid of sketching, determine the magnitude of resultant and direction using:

    • a) The Parallelogram Law.

    • b) The Triangle Rule.