AFNS Physics Chapter 02# Vectors and Equilibrium

1. Scalars and Vectors

Physical quantities can be divided into two groups based on whether they have direction or not.

1.1 Scalar Quantities

KEY CONCEPT:  A scalar quantity has only magnitude (size/number). It has NO direction.

• Mass: 5 kg (just a number — no direction needed)
• Temperature: 37°C
• Speed: 60 km/h (how fast, but not which way)
• Distance: 100 m (total path length)
• Time, Energy, Power, Work, Pressure, Density, Volume

1.2 Vector Quantities

KEY CONCEPT:  A vector quantity has both magnitude AND direction. It is represented by an arrow.

• Displacement: 5 m North (distance + direction)
• Velocity: 60 km/h East
• Force: 10 N downward
• Acceleration, Momentum, Weight, Torque, Electric Field

QUICK TIP:  Most common AFNS trick: Speed vs Velocity and Distance vs Displacement. Speed/Distance = scalar. Velocity/Displacement = vector.

1.3 Representation of Vectors

• Written as bold letter: F or with arrow on top: →F
• Drawn as an arrow: tail is starting point, head shows direction
• Length of arrow = magnitude of vector (to scale)

EXAMPLE:  A force of 20 N East is drawn as an arrow of length 2 cm pointing East (using scale: 1 cm = 10 N

2. Vector Addition — Graphical Methods

Vectors cannot be added like ordinary numbers because they have direction. We use special methods.

2.1 Triangle Law of Vector Addition

Place vectors HEAD to TAIL. The resultant is drawn from the tail of the first vector to the head of the last vector.

KEY CONCEPT:  Triangle Law: If two vectors A and B are placed head to tail, then the resultant R = A + B is the vector from the tail of A to the head of B.

EXAMPLE:  Walk 3 m East, then 4 m North. The resultant displacement = 5 m (from start to finish, at 53° from East). This uses triangle law.

2.2 Parallelogram Law of Vector Addition

Place both vectors with their TAILS at the same point. Complete a parallelogram. The diagonal from the common tail point is the resultant.

KEY CONCEPT:  Parallelogram Law: R = √(A² + B² + 2AB cosθ), where θ is the angle between vectors A and B.

EXAMPLE:  Two forces: 6 N and 8 N at 90° to each other. R = √(36 + 64 + 0) = √100 = 10 N. Direction: tan α = 8/6 → α = 53°

2.3 Important Special Cases

• Same direction (θ = 0°): R = A + B (maximum resultant)
• Opposite direction (θ = 180°): R = |A − B| (minimum resultant)
• Perpendicular (θ = 90°): R = √(A² + B²) (use Pythagoras)

QUICK TIP:  AFNS shortcut: Two equal vectors of magnitude A at angle θ → R = 2A cos(θ/2). If θ = 120°, R = 2A cos60° = 2A × 0.5 = A.

2.4 Subtraction of Vectors

A − B = A + (−B). To subtract B from A, reverse the direction of B (make it −B), then add to A.

EXAMPLE:  A = 5 m/s East, B = 3 m/s East. A − B = 5 − 3 = 2 m/s East.  |  If B was West: A − B = 5 − (−3) = 8 m/s East.

3. Vector Addition — Rectangular Components

The most powerful and precise method of vector addition. Every vector can be split into horizontal (x) and vertical (y) components.

3.1 Resolving a Vector into Components

KEY CONCEPT:  Any vector A at angle θ from the x-axis: Horizontal component Ax = A cosθ  |  Vertical component Ay = A sinθ

EXAMPLE:  A force F = 50 N at 30° above horizontal:   Fx = 50 cos30° = 50 × 0.866 = 43.3 N   |   Fy = 50 sin30° = 50 × 0.5 = 25 N

3.2 Finding the Resultant from Components

Add all x-components together and all y-components together, then combine:

• ΣFx = F1x + F2x + F3x + …
• ΣFy = F1y + F2y + F3y + …
• Resultant magnitude: R = √[(ΣFx)² + (ΣFy)²]
• Resultant direction: θ = tan⁻¹(ΣFy / ΣFx)

EXAMPLE:  Two forces: F1 = 30 N at 0° and F2 = 40 N at 90°.  ΣFx = 30 + 0 = 30 N.  ΣFy = 0 + 40 = 40 N.  R = √(900+1600) = √2500 = 50 N.  θ = tan⁻¹(40/30) = 53°

3.3 Unit Vectors

KEY CONCEPT:  Unit vector has magnitude = 1. It shows direction only. i-hat (î) = unit vector along x-axis. j-hat (ĵ) = along y-axis. k-hat (k̂) = along z-axis.

• Any vector A can be written as: A = Ax î + Ay ĵ
• Magnitude of unit vector = 1 always
• |î| = |ĵ| = |k̂| = 1

EXAMPLE:  F = 3î + 4ĵ means Fx = 3 N, Fy = 4 N. Magnitude = √(9+16) = 5 N.

QUICK TIP:  If asked ‘what is the unit vector of A = 3î + 4ĵ?’ — divide by magnitude: Â = (3î + 4ĵ)/5 = 0.6î + 0.8ĵ

4. Position Vector and Displacement

4.1 Position Vector

A position vector describes the location of a point relative to the origin. If point P has coordinates (x, y), then its position vector r = xî + yĵ.

EXAMPLE:  Point P at (3, 4): Position vector r = 3î + 4ĵ. Magnitude |r| = √(9+16) = 5 units.

4.2 Displacement Vector

KEY CONCEPT:  Displacement = Final Position − Initial Position = r₂ − r₁. It is a vector quantity showing change in position.

EXAMPLE:  Initial position r₁ = 2î + 3ĵ. Final position r₂ = 5î + 7ĵ.  Displacement = r₂ − r₁ = 3î + 4ĵ.  Magnitude = 5 units, Direction = tan⁻¹(4/3) = 53°

4.3 Distance vs Displacement

QUICK TIP:  Classic AFNS question: A person walks 3 m East then 3 m West. Distance = 6 m. Displacement = 0 m. Distance ≥ Displacement always!