A-Level — physicsFoundational4 min read

Scalars and Vectors

Distinguishing scalars from vectors; vector addition and resolution into components

Tutor AI Team
Tutor AI Team

# Scalars and Vectors — A-Level Physics

Physics quantities fall into two categories: scalars and vectors. Properly handling vectors — adding them, resolving them into components, and using them in calculations — is an essential skill that underpins virtually all A-Level Physics.

1. Scalars vs Vectors

Scalars Vectors
Have magnitude only Have magnitude and direction
Can be added normally Must be added using vector rules
Examples: mass, temperature, speed, energy, time, distance, power Examples: force, velocity, acceleration, displacement, momentum, weight

2. Representing Vectors

Vectors are represented by arrows:

  • Length = magnitude
  • Direction = direction of the arrow

Notation: F⃗\vec{F}, F\mathbf{F}, or FF with direction stated.

3. Adding Vectors

Parallel Vectors

Same direction: add magnitudes. Opposite directions: subtract magnitudes.

Non-Parallel Vectors: Triangle of Forces

Tip-to-tail method:

  1. Draw the first vector to scale
  2. Draw the second vector starting from the tip of the first
  3. The resultant is the arrow from the start of the first to the tip of the second

For three or more vectors, continue placing tip-to-tail.

Parallelogram Method

  1. Draw both vectors from the same point
  2. Complete the parallelogram
  3. The diagonal from the origin is the resultant

4. Resolving Vectors into Components

Any vector can be split into two perpendicular components (usually horizontal and vertical):

Fx=Fcos⁡θF_x = F\cos\theta Fy=Fsin⁡θF_y = F\sin\theta

Where θ\theta is the angle between the vector and the horizontal (x-axis).

Why Resolve?

Resolution simplifies 2D problems into two independent 1D problems. This is essential for:

  • Projectile motion
  • Forces on inclined planes
  • Equilibrium problems

5. Equilibrium and the Triangle of Forces

If three forces acting on a point are in equilibrium (resultant = 0), they form a closed triangle when drawn tip-to-tail.

For equilibrium: ∑Fx=0and∑Fy=0\sum F_x = 0 \quad \text{and} \quad \sum F_y = 0

6. Finding Resultant of Perpendicular Vectors

If two vectors are at right angles:

∣R⃗∣=Fx2+Fy2|\vec{R}| = \sqrt{F_x^2 + F_y^2}

θ=tan⁡−1(FyFx)\theta = \tan^{-1}\left(\frac{F_y}{F_x}\right)

Worked Example: Resolving a Force

Problem

A 50 N force acts at 30° to the horizontal. Find the horizontal and vertical components.

Fx=50cos⁡30°=50×0.866=43.3 NF_x = 50\cos30° = 50 \times 0.866 = 43.3 \text{ N} Fy=50sin⁡30°=50×0.500=25.0 NF_y = 50\sin30° = 50 \times 0.500 = 25.0 \text{ N}

Solution

Worked Example: Adding Two Perpendicular Forces

Problem

Forces of 8 N east and 6 N north act on an object. Find the resultant.

R=82+62=100=10 NR = \sqrt{8^2 + 6^2} = \sqrt{100} = 10 \text{ N} θ=tan⁡−1(6/8)=36.9° north of east\theta = \tan^{-1}(6/8) = 36.9° \text{ north of east}

Solution

Worked Example: Inclined Plane

Problem

A 20 kg block sits on a slope at 25° to the horizontal. Find the component of weight parallel to the slope and perpendicular to the slope.

W=mg=20×9.81=196.2 NW = mg = 20 \times 9.81 = 196.2 \text{ N} W∥=Wsin⁡25°=196.2×0.423=82.9 NW_{\parallel} = W\sin25° = 196.2 \times 0.423 = 82.9 \text{ N} W⊥=Wcos⁡25°=196.2×0.906=177.8 NW_{\perp} = W\cos25° = 196.2 \times 0.906 = 177.8 \text{ N}

Solution

Worked Example: Equilibrium

Problem

A lamp of weight 40 N hangs from two wires. Wire A makes 30° with the horizontal; wire B makes 45°. Find the tension in each wire.

Horizontal equilibrium: TAcos⁡30°=TBcos⁡45°T_A\cos30° = T_B\cos45°

Vertical equilibrium: TAsin⁡30°+TBsin⁡45°=40T_A\sin30° + T_B\sin45° = 40

From horizontal: TA=TB×cos⁡45°/cos⁡30°=TB×0.707/0.866=0.816TBT_A = T_B \times \cos45°/\cos30° = T_B \times 0.707/0.866 = 0.816T_B

Substitute into vertical: 0.816TB×0.5+TB×0.707=400.816T_B \times 0.5 + T_B \times 0.707 = 40 0.408TB+0.707TB=400.408T_B + 0.707T_B = 40 1.115TB=401.115T_B = 40 TB=35.9T_B = 35.9 N, TA=29.3T_A = 29.3 N

Solution

8. Practice Questions

    1. A velocity of 12 m/s acts at 40° above the horizontal. Calculate the horizontal and vertical components. (2 marks)
    1. Two forces of 5 N and 12 N act at right angles. Find the magnitude and direction of the resultant. (3 marks)
    1. A 10 kg block is on a 35° slope. Calculate the component of weight (a) parallel to the slope, (b) perpendicular to the slope. (4 marks)
    1. Three forces act on a point: 10 N north, 7 N east, and an unknown force FF. The system is in equilibrium. Calculate FF. (4 marks)

    Answers

    1. vx=12cos⁡40°=9.19v_x = 12\cos40° = 9.19 m/s, vy=12sin⁡40°=7.71v_y = 12\sin40° = 7.71 m/s.

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Frequently Asked Questions

When do I use sin and when cos for resolution?

If θ\theta is measured from the horizontal: horizontal component = Fcos⁡θF\cos\theta, vertical = Fsin⁡θF\sin\theta. If θ\theta is from the vertical, swap them. Always draw a diagram.

Can a scalar ever be negative?

Scalars like temperature can be negative, but most (mass, speed, distance) cannot. However, the component of a vector can be negative, indicating direction.

Summary

  • Scalars: magnitude only; Vectors: magnitude + direction
  • Add vectors using tip-to-tail or parallelogram methods
  • Resolve vectors: Fx=Fcos⁡θF_x = F\cos\theta, Fy=Fsin⁡θF_y = F\sin\theta
  • Resultant of perpendicular vectors: Pythagoras + trigonometry
  • Equilibrium: ∑Fx=0\sum F_x = 0, ∑Fy=0\sum F_y = 0
Tags:
#a-level#physics#mechanics#scalars#vectors#components

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