Vectors and Scalars

Vectors
Scalars and Vectors

  • In physics, two types of quantities exist:

    1. Scalars

    • Defined as quantities that possess only a magnitude, without any spatial direction.

    • Examples of Scalars:

      • Distance

      • Speed

      • Area

      • Volume

      • Work

      • Energy

      • Temperature

    1. Vectors

    • Defined as quantities that possess both a magnitude and a spatial direction.

    • Examples of Vectors:

      • Displacement

      • Velocity

      • Acceleration

      • Force

      • Torque

Representing Vectors

  • A vector is visually represented by an arrow:

    • The length of the arrow corresponds to its magnitude.

    • The orientation of the arrow indicates its direction.

  • Notation for vectors:

    • A vector can be denoted by a letter with an arrow on top, e.g., $oldsymbol{ ilde{A}}$ or $ar{A}$.

    • The magnitude of the vector $oldsymbol{ ilde{A}}$ or $ar{A}$ is denoted by $|oldsymbol{ ilde{A}}|$ or simply $A$.

    • Many textbooks utilize boldface to represent vectors, e.g., $ extbf{A}$.

    • In this course, both notations for vectors will be applied.

Equal Vectors and Negative Vectors

  • Equal Vectors:

    • Two vectors are considered equal if they have both the same magnitude and the same direction.

    • Example: If $oldsymbol{ ilde{A}} = oldsymbol{ ilde{B}}$, then their magnitudes and directions match.

  • Negative Vectors:

    • The negative of a vector $oldsymbol{ ilde{A}}$ is denoted as $-oldsymbol{ ilde{A}}$.

    • This vector has the same magnitude as $oldsymbol{ ilde{A}}$ but is directed opposite to $oldsymbol{ ilde{A}}$.

Unit Vectors

  • Unit Vectors:

    • Defined as vectors with a magnitude of one unit.

    • Notation: Represented by a hat (^) on top of the letter.

    • Example: A unit vector in the direction of $oldsymbol{ ilde{A}}$ is given by $oldsymbol{ ilde{a}} = rac{oldsymbol{ ilde{A}}}{|oldsymbol{ ilde{A}}|}$.

    • Unit vectors help establish a particular direction.

Coordinate System

  • A coordinate system enables unique definition of the position of a point in space.

    • Example: The Cartesian Coordinate System is formed using three mutually perpendicular lines, each with a positive and negative direction.

  • Question: Who decides on a coordinate system?

Examples of Coordinate Systems

  • Geographical Coordinate Systems:

    • Directions: East, West, North, South

  • **Spherical Coordinate Systems

  • Polar Coordinate System

  • Cylindrical Coordinate Systems**

Reference Frame

  • A reference frame consists of physical reference points in which a coordinate system is uniquely located and oriented.

  • Examples of Reference Frames:

    • The Earth as a Fixed Frame of Reference.

    • The Human Body as a Reference Frame (not fixed).

Human Body as a Reference

  • Frontal or Coronal Plane:

    • Divides the body into front and back halves; involves side-to-side movements.

  • Sagittal or Lateral Plane:

    • Divides the body into left and right halves; involves forward and backward movements.

  • Horizontal or Axial/Transverse Plane:

    • Divides the body into top and bottom halves; involves twisting movements.

Describing Vectors: Magnitude and Direction
Method 1

  • To describe vectors, it is essential to measure their direction concerning a reference direction.

    • In 2-D space, select two perpendicular directions and establish them as the $x-y$ coordinate axes.

    • In relation to the aforementioned axes, a vector $oldsymbol{ ilde{A}}$ can be described as:

    • $oldsymbol{ ilde{A}} ext{ is defined as } (A, heta)$ where $A$ is the magnitude, and $ heta$ is the angle with respect to the positive x-axis.

Method 2: Vector Components

  1. Components of Vectors:

    • Vector $oldsymbol{ ilde{A}}$ can be described in terms of the coordinates $Ax, Ay$ of the tip of the vector $oldsymbol{ ilde{R}}$.

  2. Definitions of Components:

    • $A_x$: Represents how much $oldsymbol{ ilde{A}}$ extends in the $x$-direction. Known as the projection of $oldsymbol{ ilde{A}}$ on the $x$-axis.

    • $A_y$: Represents how much $oldsymbol{ ilde{A}}$ extends in the $y$-direction. Known as the projection of $oldsymbol{ ilde{A}}$ on the $y$-axis.

Finding Vector Components from Magnitude and Direction

  • Given $oldsymbol{ ilde{A}} ext{ as } (A, heta)$:

  1. To find $A_x$:

    • Drop a perpendicular from the tip of $oldsymbol{ ilde{A}}$ (point $R$) onto the $x$-axis.

    • The intersection point $P$ is at a distance $A_x$ from the origin $O$.

    • From the right triangle $ riangle OPR$, the relationship is defined as follows:
      extcos(heta)=OPOR=AxAext{cos}( heta) = \frac{OP}{OR} = \frac{A_x}{A}

    • Therefore, the equation becomes:
      Ax=A⋅extcos(heta)A_x = A \cdot ext{cos}( heta)

  2. To find $A_y$:

    • Drop a perpendicular onto the $y$-axis.

    • Point $Q$ intersects the $y$-axis at a distance $A_y$ from the origin $O$.

    • From the right triangle $ riangle OQR$, the relationship is defined as follows:
      extcos(90−heta)=OQOR=AyAext{cos}(90 - heta) = \frac{OQ}{OR} = \frac{A_y}{A}

    • Trigonometric identity states:
      extcos(90−heta)=extsin(heta)ext{cos}(90 - heta) = ext{sin}( heta)

    • Thus it follows:
      extsin(heta)=AyAext{sin}( heta) = \frac{A_y}{A}

    • Therefore,
      Ay=A⋅extsin(heta)A_y = A \cdot ext{sin}( heta)

  • -

Summary of Components:

  • Ax=A⋅extcos(heta)A_x = A \cdot ext{cos}( heta)

  • Ay=A⋅extsin(heta)A_y = A \cdot ext{sin}( heta)

  • Note: Components $Ax$ and $Ay$ can be positive or negative depending on which quadrant the vector is located. The angle $ heta$ is measured counter-clockwise from the positive $x$-axis.

Example Problem: Force Calculation

  • Given Problem: A force of 6 N acts at an angle of 60° north of west.

  • Objective: Determine the effective forces acting along the west and north directions.

Finding Magnitude and Direction from Components

  • Given $Ax$ and $Ay$, the overall magnitude $A$ and the angle $ heta$ can be calculated as follows:

  1. Using the Pythagorean Theorem:

    • In right triangle $ riangle OPR$:
      OP2+PR2=OR2OP^2 + PR^2 = OR^2

    • Substituting in the components leads to:
      A<em>x2+A</em>y2=A2A<em>x^2 + A</em>y^2 = A^2

    • Thus, the magnitude is defined as:
      A=A<em>x2+A</em>y2A = \sqrt{A<em>x^2 + A</em>y^2}

  2. Finding Angle b8:

    • Using the trigonometric ratio in triangle $ riangle OPR$ gives:
      tan⁡(heta)=PROP\tan( heta) = \frac{PR}{OP}

    • Hence, substituting the components:
      tan⁡(heta)=A<em>yA</em>x\tan( heta) = \frac{A<em>y}{A</em>x}

    • Therefore,
      heta=tan⁡−1(A<em>yA</em>x)heta = \tan^{-1}(\frac{A<em>y}{A</em>x})

    • Note: The calculator will yield a value restricted to b8 from −90° ≤ b8 ≤ 90°.

Considerations for Non-Acute Angles

  • Observe how the derived formulas alter when b8 is not acute:

  • Quadrant 1:

    • $Ax > 0$, $Ay > 0$

  • Quadrant 2:

    • $Ax < 0$, $Ay > 0$

  • Quadrant 3:

    • $Ax < 0$, $Ay < 0$

  • Quadrant 4:

    • $Ax > 0$, $Ay < 0$

  • Note: The coordinates (x,y) and (−x,−y) yield the same tangent function. For instance, (1,1) and (-1,-1) both have the same value of tan b8 = 1.

  • Each value of sin, cos, and tan corresponds to two angles. It is crucial to select the appropriate angle based on the signs of $x$ and $y$ coordinates.

Calculating Angle using Inverse Functions
Given Range for Inverse Functions:

  • For b8:

    • ext{sin}^{-1}: ext{Range}: -90° ≤ b8 ≤ 90°

    • ext{cos}^{-1}: ext{Range}: 0° ≤ b8 ≤ 180°

    • ext{tan}^{-1}: ext{Range}: -90° ≤ b8 ≤ 90°

  • When utilizing inverse trigonometric functions, the calculator provides only one value.

Angle Correction:

  • You may need to add corrective angles of 180° or 360° depending on the quadrant location of the point.

Example: Angle with the Positive x-Axis for a Vector in the Second Quadrant

  • Consider the vector $oldsymbol{ ilde{A}}$ with components (-3, 2):

  • To calculate the angle it makes with the positive x-axis:

    • Start with:
      heta=tan⁡−1(A<em>yA</em>x)=tan⁡−1(−23)=−33.7°heta = \tan^{-1}(\frac{A<em>y}{A</em>x}) = \tan^{-1}(\frac{-2}{3}) = -33.7°

    • Since this angle corresponds to the fourth quadrant (add 180°):
      extFinalAngle=−33.7°+180°=146.3°ext{Final Angle} = -33.7° + 180° = 146.3°

Clockwise Angle Calculation with the Positive x-Axis

  • Vector in 1st or 4th Quadrant:

    • Calculate as $ heta = \tan^{-1}(\frac{Ay}{Ax})$

  • Vector in 2nd or 3rd Quadrant:

    • Calculate as $ heta = \tan^{-1}(\frac{Ay}{Ax}) + 180°$

Example Problem: Vector (4, -3)

  • Task: Draw the vector and find its magnitude and direction.

Adding Vectors

  • Discuss two main methods for vector addition:

    1. Graphical Method

    2. Adding Vectors Using Components

Graphical Method for Adding Vectors

  • When hiking on a trail, two paths $oldsymbol{ ilde{A}}$ and $oldsymbol{ ilde{B}}$ can be analyzed:

    • Determine the resultant displacement $oldsymbol{ ilde{C}}$.

    • Resultant displacement is defined as starting at the tail of $oldsymbol{ ilde{A}}$ and ending at the head of $oldsymbol{ ilde{B}}$:

    • ildeA+ildeB=ildeC\boldsymbol{ ilde{A}} + \boldsymbol{ ilde{B}} = \boldsymbol{ ilde{C}}

    • This operation, known as vector addition, produces the sum of the vectors.

    • Geometrically, ildeC\boldsymbol{ ilde{C}} is the third side of a triangle formed by $oldsymbol{ ilde{A}}$ and $oldsymbol{ ilde{B}}$.

Head-to-Tail Method

  • Vectors are arranged in a manner where the tail of one vector is placed at the head of the previous vector:

    • The order of placement does not affect the resultant.

    • The resultant vector extends from the tail of the first vector to the head of the last vector.

Vector Addition Example: Finding Resultant Force

  • Example Problem: Identify the force acting on a box using the head-to-tail method.

Evaluating Vector Addition: Head-to-Tail Method

  • Step 1: Position vectors by arranging the tail of one at the head of the preceding vector.

  • Investigate if the resultant matches previous calculations.

Vector Subtraction

  • Vector subtraction is expressed as:

    • ildeA−ildeB=ildeA+(−ildeB)\boldsymbol{ ilde{A}} - \boldsymbol{ ilde{B}} = \boldsymbol{ ilde{A}} + (-\boldsymbol{ ilde{B}})

  • This signifies vector addition of the first vector and the reverse of the second vector.

Practical Problem Solving
Example: Identifying Correct Representation of Vector Addition

  • Analyze diagrams varied to identify which figure correctly represents as the resultant vector for $oldsymbol{ ilde{A}} + oldsymbol{ ilde{B}}$.

Components of Vector Addition

  • The additive operation is derived as follows:

    • ildeA+ildeB=ildeC\boldsymbol{ ilde{A}} + \boldsymbol{ ilde{B}} = \boldsymbol{ ilde{C}}

    • Let components of $oldsymbol{ ilde{A}} ext{ be } (Ax, Ay)$ and components of $oldsymbol{ ilde{B}} ext{ be } (Bx, By)$:

    • The components of $oldsymbol{ ilde{C}}$ are established as:

      • C<em>x=A</em>x+BxC<em>x = A</em>x + B_x

      • C<em>y=A</em>y+ByC<em>y = A</em>y + B_y

  • Therefore, using these components, the overall magnitude $C$ and angle $ heta$ can be calculated as:

    • C=C<em>x2+C</em>y2C = \sqrt{C<em>x^2 + C</em>y^2}

    • heta=tan⁡−1(C<em>yC</em>x)heta = \tan^{-1}(\frac{C<em>y}{C</em>x})

Unit Vector Notation

  • Define unit vectors for the $x$, $y$, and $z$ directions as $oldsymbol{ ilde{i}}$, $oldsymbol{ ilde{j}}$, $oldsymbol{ ilde{k}}$, respectively.

  • For vector components, the representation forms:

    • ildeA=A<em>xildei+A</em>yildej\boldsymbol{ ilde{A}} = A<em>x \boldsymbol{ ilde{i}} + A</em>y \boldsymbol{ ilde{j}}

  • Magnitude interpretation as per the sign of components.

Adding Vectors in Unit Vector Notation

  • To add vectors using unit vector notation, sum the components:

    • ildeA=A<em>xildei+A</em>yildej\boldsymbol{ ilde{A}} = A<em>x \boldsymbol{ ilde{i}} + A</em>y \boldsymbol{ ilde{j}}

    • ildeB=B<em>xildei+B</em>yildej\boldsymbol{ ilde{B}} = B<em>x \boldsymbol{ ilde{i}} + B</em>y \boldsymbol{ ilde{j}}

    • As a result,
      ildeC=(A<em>x+B</em>x)ildei+(A<em>y+B</em>y)ildej\boldsymbol{ ilde{C}} = (A<em>x + B</em>x) \boldsymbol{ ilde{i}} + (A<em>y + B</em>y) \boldsymbol{ ilde{j}}

Types of Vector Multiplication

  • There are three forms of multiplicative operations with vectors:

    1. Multiplying a vector by a scalar resulting in another vector.

    2. Multiplying two vectors yielding a scalar (Dot Product).

    3. Multiplying two vectors resulting in a vector (Cross Product).

Multiplying Vectors by Scalars

  • Multiplying a vector by a scalar produces a new vector:

    • Magnitude: Equal to the product of the original vector's magnitude and the scalar.

    • If the scalar is positive, the new vector aligns in the same direction as the original vector.

    • If the scalar is negative, the new vector points in the opposite direction.

Example Problem: Martin's Displacement

  • Martin's Journey:

  1. Walks 2 km at an angle of 45° east of south (Leg 1).

  2. Changes direction and walks 3 km at 60° north of east (Leg 2). Tasks: A. Determine Martin’s displacement (both magnitude and direction using graphical method). B. Calculate his displacement using the analytical method.

    • Thought Process: Identify given quantities, unknowns, and the essential values to be found for the resolution of the problem.