Example (1): Two perpendicular forces

• Two forces act at one point a force of 3 N to the east and a force of 4 N to the north. Find their resultant.
• Solution:

1. Take east as the positive horizontal direction and north as the positive vertical direction.
2. ΣX = 3 + 0 = 3 N, and ΣY = 0 + 4 = 4 N.
3. R = √(3² + 4²) = √25 = 5 N.
4. tan θ = 4 ÷ 3, so θ ≈ 53.13° north of east.

Example (2): Forces in more than one quadrant

• Three forces act at one point 6 N in the direction of the positive x-axis, 8 N in the direction of the positive y-axis, and 10 N making an angle of 180° with the positive x-axis. Find the resultant.
• Solution:

1. ΣX = 6 + 0 + 10 × cos 180° = 6 − 10 = −4 N.
2. ΣY = 0 + 8 + 10 × sin 180° = 8 + 0 = 8 N.
3. R = √((−4)² + 8²) = √80 = 4√5 ≈ 8.94 N.
4. tan θ = 8 ÷ (−4) = −2. The resultant lies in the second quadrant, so θ ≈ 116.57° from the positive x-axis.

Example (3): Using the special angles

• Three forces act at one point 10 N along the x-axis, 6 N making an angle of 60° with the same axis, and 6 N making an angle of 180° with it. Find the magnitude and the direction of the resultant.
• Solution:

1. ΣX = 10 + 6 × cos 60° + 6 × cos 180° = 10 + 3 − 6 = 7 N.
2. ΣY = 0 + 6 × sin 60° + 0 = 6 × √3 ÷ 2 = 3√3 ≈ 5.20 N.
3. R = √(7² + (3√3)²) = √(49 + 27) = √76 = 2√19 ≈ 8.72 N.
4. tan θ = 3√3 ÷ 7 ≈ 0.742, so θ ≈ 36.59°.

Example (4): Three forces of equal magnitude

• Three forces, each of magnitude 8 N, act at one point and make angles of 30°, 90° and 150° with the positive x-axis respectively.
• Find the resultant.
• Solution:

1. ΣX = 8 × cos 30° + 8 × cos 90° + 8 × cos 150° = 8 × √3 ÷ 2 + 0 − 8 × √3 ÷ 2 = 0.
2. ΣY = 8 × sin 30° + 8 × sin 90° + 8 × sin 150° = 4 + 8 + 4 = 16 N.
3. The resultant is 16 N, acting vertically upwards (in the direction of 90°).