Force Equilibrium Example
Find the resultant force and determine if a particle is in equilibrium under three concurrent forces.
Problem
Force Diagram
Arrows represent force vectors. Magnitude and direction shown for each force.
A particle is subjected to three concurrent forces:
- F1 = 10 kN at 0° (horizontal right)
- F2 = 8 kN at 90° (vertical up)
- F3 = 6 kN at 180° (horizontal left)
Find:
- The magnitude and direction of the resultant force
- Determine if the particle is in equilibrium
Given
- Force F₁
- 10 kN at 0°
- Force F₂
- 8 kN at 90°
- Force F₃
- 6 kN at 180°
Solution
Step 1: Resolve forces into components
Break each force into x and y components:
F1x = 10 cos(0°) = 10 kN → , F1y = 10 sin(0°) = 0 kN
F2x = 8 cos(90°) = 0 kN , F2y = 8 sin(90°) = 8 kN ↑
F3x = 6 cos(180°) = -6 kN ← , F3y = 6 sin(180°) = 0 kN
Step 2: Sum the components
Sum all x-components and y-components separately:
ΣFₓ = 10 + 0 + (-6) = 4 kN →
ΣFᵧ = 0 + 8 + 0 = 8 kN ↑
Step 3: Calculate resultant magnitude and direction
Use Pythagorean theorem and inverse tangent:
R = √(ΣFₓ² + ΣFᵧ²) = √(4² + 8²) = √(16 + 64) = √80
R = 8.94 kN
θ = tan⁻¹(ΣFᵧ / ΣFₓ) = tan⁻¹(8/4) = tan⁻¹(2)
θ = 63.4° (above horizontal)
Step 4: Check equilibrium
For equilibrium, the resultant must be zero:
R = 8.94 kN ≠ 0
The particle is NOT in equilibrium.
Final Answer
- • Resultant force R = 8.94 kN at 63.4° above horizontal
- • The particle is NOT in equilibrium (resultant ≠ 0)
Practice Problem
Try solving this similar problem:
Three forces act on a particle: 12 kN at 0°, 9 kN at 120°, and 15 kN at 240°. Find the resultant force.
Show Answer
ΣFₓ = 12 + 9cos(120°) + 15cos(240°) = 0 kN
ΣFᵧ = 0 + 9sin(120°) + 15sin(240°) = 0 kN
R = 0 kN - The particle IS in equilibrium!