6  Vectors in Two Dimensions

6.1 Syllabus inquiry question

  • How is the motion of an object moving in a straight line described and predicted?
Feynman Insight

From The Feynman Lectures on Physics, Vol I, Chapter 11:

Vector addition is not a trick; it is a statement about how nature combines directions. Treating velocities like scalars loses the geometry of motion.

6.2 Learning Objectives

  • Represent vectors using components and diagrams.
  • Resolve vectors into perpendicular components.
  • Add and subtract vectors in two dimensions.
  • Apply vector methods to velocity and displacement.

6.3 Content

6.3.1 Vector components

A vector \(\vec{R}\) at angle \(\theta\) from the horizontal has components:

\[R_x = R\cos\theta, \quad R_y = R\sin\theta\]

Tip

The angle is always measured from the positive x-axis (east direction) unless otherwise specified.

6.3.2 Interactive: Vector Resolution

The diagram below shows a vector being resolved into its x and y components:

6.3.3 Vector addition

Graphical methods (head-to-tail) and component methods both yield the resultant. The component method is preferred for calculations.

6.3.4 Interactive: Vector Addition (Tail-to-Head)

Add two vectors using the tail-to-head method. The resultant (dashed) connects the start to the end.

Method using components:

  1. Resolve each vector into x and y components
  2. Add all x components: \(R_x = A_x + B_x\)
  3. Add all y components: \(R_y = A_y + B_y\)
  4. Find magnitude: \(R = \sqrt{R_x^2 + R_y^2}\)
  5. Find direction: \(\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right)\)

6.3.5 Resultant magnitude and direction

\[R = \sqrt{R_x^2 + R_y^2}, \quad \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right)\]

Warning

Watch the quadrant! If \(R_x < 0\), the angle from \(\tan^{-1}\) needs adjustment (add 180°).

6.3.6 Interactive: Three-Vector Addition

When adding more than two vectors, the component method becomes essential:

6.4 Worked Examples

6.4.1 Example 1: Resolve a vector

A displacement of 50 m is directed 30° north of east.

Solution:

  1. \(R_x = 50\cos(30°) = 50 \times 0.866 = 43.3\) m (east)
  2. \(R_y = 50\sin(30°) = 50 \times 0.5 = 25.0\) m (north)
  3. Components describe the east and north parts of the motion.

6.4.2 Example 2: Add two vectors

A walker goes 3.0 m east, then 4.0 m north.

Solution:

  1. Components: \(R_x = 3.0\) m, \(R_y = 4.0\) m
  2. Resultant magnitude: \(R = \sqrt{3.0^2 + 4.0^2} = \sqrt{25} = 5.0\) m
  3. Direction: \(\theta = \tan^{-1}(4.0/3.0) = 53°\) north of east

6.4.3 Example 3: Subtract vectors (relative motion)

A boat’s velocity is 6.0 m/s east. The current is 2.0 m/s north. Find the boat’s velocity relative to the water.

Solution:

  1. \(\vec{v}_{bw} = \vec{v}_{be} - \vec{v}_{we}\)
  2. Components: \(v_x = 6.0\) m/s, \(v_y = -2.0\) m/s (subtract current)
  3. Speed: \(\sqrt{6.0^2 + 2.0^2} = 6.3\) m/s
  4. Direction: \(\tan^{-1}(2.0/6.0) = 18°\) south of east

6.4.4 Example 4: Find components from magnitude and direction

A force of 25 N acts at 60° above the horizontal.

Solution:

  1. \(F_x = 25\cos(60°) = 12.5\) N (horizontal)
  2. \(F_y = 25\sin(60°) = 21.7\) N (vertical)

6.5 Common Misconceptions

Common Misconceptions

  • Misconception: Components are always equal to the magnitude. Correction: Components depend on direction; only at 45° are they equal.

  • Misconception: Vector addition is the same as adding magnitudes. Correction: Directions change the result; 3 m east + 4 m north ≠ 7 m.

  • Misconception: A negative component means the vector is negative. Correction: It only indicates direction along the axis.

  • Misconception: \(\tan^{-1}\) always gives the correct angle. Correction: You must check the quadrant based on component signs.

6.6 Practice Questions

6.6.1 Easy (2 marks)

Resolve a 10 m displacement at 60° above the horizontal into components.

  • Correct cosine and sine components (2)

Answer: - \(x = 10\cos(60°) = 5.0\) m - \(y = 10\sin(60°) = 8.7\) m

6.6.2 Medium (4 marks)

Two forces act on a point: 8 N east and 6 N north. Find the resultant magnitude and direction.

  • Correct components and resultant magnitude (2)
  • Correct direction (2)

Answer: - \(R = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\) N - \(\theta = \tan^{-1}(6/8) = 37°\) north of east

6.6.3 Hard (5 marks)

An aircraft travels 200 km/h north relative to the air. A wind of 60 km/h blows east. Find the ground velocity and its direction.

  • Correct vector model (1)
  • Components and magnitude (2)
  • Correct direction statement (2)

Solution:

The ground velocity is the sum of air velocity and wind velocity:

  • \(v_x = 0 + 60 = 60\) km/h (east)
  • \(v_y = 200 + 0 = 200\) km/h (north)

Ground speed: - \(v = \sqrt{60^2 + 200^2} = \sqrt{43600} = 209\) km/h

Direction: - \(\theta = \tan^{-1}(60/200) = 17°\) east of north

Answer: 209 km/h at 17° east of north

6.7 Multiple Choice Questions

Test your understanding with these interactive questions:

6.8 Summary

Key Takeaways
  • Components encode vector direction in x and y axes.
  • Vector addition can be solved with components and Pythagoras.
  • Direction is found using inverse tangent with correct quadrant.
  • Vector methods underpin two-dimensional motion.

6.9 Self-Assessment

Check your understanding:

After studying this section, you should be able to: