Example (1): Finding an unknown force that keeps the body in equilibrium

• Three forces act on a body at rest 15 N in the direction of the positive x-axis, 20 N in the direction of the positive y-axis, and a third force F. Find the magnitude and the direction of F so that the body is in equilibrium.
• Solution:

1. Equilibrium requires ΣX = 0, so 15 + F_x = 0, giving F_x = −15 N.
2. ΣY = 0, so 20 + F_y = 0, giving F_y = −20 N.
3. Magnitude of F = √((−15)² + (−20)²) = √625 = 25 N.
4. tan θ = (−20) ÷ (−15) = 4 ÷ 3. The force lies in the third quadrant, so θ ≈ 180° + 53.13° = 233.13° from the positive x-axis.

Example (2): A body hung by two strings

• A body of weight 20 N is hung from one point by two strings.
• The left string makes an angle of 60° with the horizontal, and the right string makes an angle of 30° with the horizontal.
• Find the tension in each string when the body is in equilibrium.
• Solution:

Example (3): Three forces forming a right-angled triangle

• Solution:

1. The third force must satisfy ΣX = 0 and ΣY = 0, so its components are (−5, −12).
2. Its magnitude is √(5² + 12²) = √169 = 13 N.
3. tan θ = (−12) ÷ (−5) = 12 ÷ 5, and the force lies in the third quadrant, so θ ≈ 180° + 67.38° = 247.38°.
4. The three forces form a right-angled triangle with sides 5, 12 and 13, as expected.

Example (4): A body hung by a string and pulled sideways

• A body of weight 40 N hangs from a vertical string.
• A horizontal force of 30 N pulls it sideways so that it is in equilibrium in an inclined position.
• Find the tension in the string and the angle the string makes with the vertical.
• Solution:

1. Let the tension be T and the angle with the vertical be θ. Then T sin θ = 30 (horizontal) and T cos θ = 40 (vertical).
2. T² = 30² + 40² = 900 + 1600 = 2500, so T = 50 N.
3. tan θ = 30 ÷ 40 = 3 ÷ 4, so θ ≈ 36.87°.
4. 50 × 0.6 = 30 and 50 × 0.8 = 40, which is correct.