Vectors – Components, Addition, and Trigonometry

Scalar and Vector Quantities

• Scalar Quantity

  • Has magnitude only, no direction.

  • Common examples mentioned:

  • Mass

  • Temperature

  • Work (in physics sense, when treated as scalar product)

  • Time

  • Distance

  • Speed

  • Measurement concerns “how much” only.

• Vector Quantity

  • Describes motion or influence with both magnitude and direction.

  • Examples explicitly listed:

  • Weight

  • Force

  • Velocity

  • Displacement

  • Acceleration

  • Momentum

  • Torque

  • Formal definition: “movement of a body with respect to its magnitude and direction.”

  • Visual representation: arrow whose length encodes magnitude and whose orientation encodes direction.

  • Notation tip: boldface $\mathbf{v}$, letter with arrow $\vec{v}$ or coordinate pair $(v<em>x, v</em>y)$ .

Cartesian Plane & Direction Conventions

• Standard $x$–$y$ plane sketched with grid from $-10$ to $10$ on each axis.

• Four quadrants permit representation of any 2-D vector.

• Cardinal directions overlaid for navigation-style wording:

  • Positive $x$ axis → East (E)

  • Positive $y$ axis → North (N)

  • Negative $x$ axis → West (W)

  • Negative $y$ axis → South (S)

• “North of East” phrasing = start at East and rotate counter-clockwise toward North.

  • Example diagram showed a vector labeled “N of E” drawn into first quadrant.

Right-Triangle Foundation for Vectors

• Any 2-D vector can be treated as the hypotenuse of a right triangle whose legs are its components.

  • Adjacent leg = horizontal ($x$-component)

  • Opposite leg = vertical ($y$-component)

  • Angle $\theta$ measured from the positive $x$ axis unless problem states otherwise.

• Naming Recap (slide used color blocks):

  • Y Component → vertical direction of a vector.

  • X Component → horizontal direction of a vector.

  • Resultant Vector ($R$) → sum of $x$ and $y$ components; in one-vector context it is simply the original vector.

Importance of Vector Components

• Simplification of 2-D Vectors: Any 2-D vector can be treated as the hypotenuse of a right triangle, and its legs are its horizontal ($x$-component) and vertical ($y$-component) parts. This allows complex 2-D problems to be broken into simpler 1-D problems along each axis.

• Mathematical Operations: Components make it easy to add or subtract multiple vectors analytically. You simply sum all the $x$-components to get the resultant $x$-component ($R<em>x$) and all the $y$-components to get the resultant $y$-component ($R</em>y$).

• Application of 1-D Equations: Once a vector is resolved into its components, you can apply familiar 1-D kinematic or dynamic equations to each component independently.

• Practicality: This method is critical in fields like engineering, where accurate calculations of force components, for example, are essential to ensure the stability and safety of structures or systems.

Right-Triangle Relations and Pythagorean Theorem

• Key right-triangle relations (SOH-CAH-TOA):

  • $\sin\theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}}$

  • $\cos\theta = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}$

  • $\tan\theta = \dfrac{\text{Opposite}}{\text{Adjacent}}$

• Pythagorean Theorem for magnitude recovery:

  • $R = \sqrt{(R<em>x)^2 + (R</em>y)^2}$

Mathematical Tools Employed

• Trigonometric Functions (sine, cosine, tangent) to resolve or combine components.

• Pythagorean Theorem for resultant magnitude.

• Head-to-Tail Method for adding multiple vectors graphically.

• Component (analytical) method for adding vectors numerically.

Worked Example 1: Single Displacement Vector

Vector: $F = 600\,\text{m}$, $\theta = 50^{\circ}$ “North of East”

1. Break into components

   • $F_x = F \cos 50^{\circ} = (600\,\text{m}) (\cos 50^{\circ}) = 385.67\,\text{m}$

   • $F_y = F \sin 50^{\circ} = (600\,\text{m}) (\sin 50^{\circ}) = 459.63\,\text{m}$

2. Verification by Pythagorean Theorem

   • $\sqrt{(385.67\,\text{m})^2 + (459.63\,\text{m})^2} = \sqrt{360001.04\,\text{m}^2} \approx 600\,\text{m}$ (matches original magnitude).

3. Interpretation

   • Displacement of $459.63\,\text{m}$ North and $385.67\,\text{m}$ East enters QI.

   • Arrow would be labelled “$600\,\text{m}, 50^{\circ} \text{ N of E}$.”

Worked Example 2: Addition of Two Forces (Vector Sum)

Given forces directed into the South of West (SoW) quadrant:

1. $\mathbf{F}<em>1 = 25\,\text N,\; \theta</em>1 = 50^{\circ}\, \text{SoW}$

2. $\mathbf{F}<em>2 = 40\,\text N,\; \theta</em>2 = 15^{\circ}\, \text{SoW}$

Assume angles measured from the West axis toward South.

Step 1 – Resolve Each Force

• Sign convention: West = negative $x$, South = negative $y$.

Force 1 ($F_1$):

• $F<em>{1x} = F</em>1 \cos 50^{\circ} = (25\,\text N)(\cos 50^{\circ}) = 16.07\,\text N \text{ (pointing West } \Rightarrow \text{ negative)}$

• $F<em>{1y} = F</em>1 \sin 50^{\circ} = (25\,\text N)(\sin 50^{\circ}) = 19.15\,\text N \text{ (pointing South } \Rightarrow \text{ negative)}$

Force 2 ($F_2$):

• $F<em>{2x} = F</em>2 \cos 15^{\circ} = (40\,\text N)(\cos 15^{\circ}) = 38.64\,\text N \text{ (West, negative)}$

• $F<em>{2y} = F</em>2 \sin 15^{\circ} = (40\,\text N)(\sin 15^{\circ}) = 10.35\,\text N \text{ (South, negative)}$

Step 2 – Sum Components

• $F_{\text{total }x} = -(16.07 + 38.64)\,\text N = -54.71\,\text N$

• $F_{\text{total }y} = -(19.15 + 10.35)\,\text N = -29.50\,\text N$

(negative signs confirm resultant lies in South-West quadrant.)

Step 3 – Resultant Magnitude

• $F_R = \sqrt{(-54.71\,\text N)^2 + (-29.50\,\text N)^2} = \sqrt{2993.18\,\text N^2} \approx 62.16\,\text N$

Step 4 – Resultant Direction

• Use tangent with absolute component values:

  • $\theta = \tan^{-1}\left(\dfrac{|F<em>y|}{|F</em>x|}\right) = \tan^{-1}\left(\dfrac{29.50}{54.71}\right) \approx 28.33^{\circ}$

• Phrase: $28.3^{\circ}$ South of West OR equivalently $62.16\,\text N$ at $208.3^{\circ}$ standard position.

Step 5 – Graphical Head-to-Tail Check

• Original slides sketched both arrows end-to-end; the resulting diagonal matched computed $62.16\,\text N$ vector, validating arithmetic method.

General Problem-Solving Strategy for 2-D Vectors

• Draw a clear diagram with a labelled Cartesian grid.

• Mark given magnitudes and angles; note quadrant signs.

• Use SOH-CAH-TOA to generate components.

• Treat components algebraically (include signs) when adding or subtracting vectors.

• Compute resultant magnitude with $R = \sqrt{R<em>x^2 + R</em>y^2}$.

• Find direction with inverse tangent; adjust by quadrant rules.

• State answer with both magnitude and a direction phrase (e.g.-

  • “$R = 5.2\,\text m$ at $35^{\circ}$ N of E” or

  • “$R = 5.2\,\text m$, $\theta = 35^{\circ}$ measured from $+x$ axis.”).

Practical & Pedagogical Connections

• Content builds on earlier lectures on basic kinematics & trigonometry.

• Vectors underpin General Physics I topics: equilibrium of forces, projectile motion, momentum conservation.

• Vector decomposition avoids memorizing case-specific formulas; once components are found, one can apply familiar 1-D equations along each axis.

• Ethical/Practical note: Accurate component work is critical in engineering—e.g., mis-calculating a force component could under-design a structural beam.

Summary Cheat-Sheet

• Identify scalar vs vector.

• Draw arrow, pick origin, note angle.

• Resolve: $x = F\cos\theta$, $y = F\sin\theta$.

• Add components when combining vectors.

• Recover magnitude & angle.

• Express answer with proper units and directional language.