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EQUILIBRIUM OF FORCES
Equilibrium of Forces — Edwin Ogie Library
Comprehensive e-book: static & dynamic equilibrium, resultant & equilibrant forces, torque, moments, worked examples and a CBT quiz for JSS/Senior students.
Introduction
An object is in equilibrium if it experiences no acceleration — i.e. the net force and net moment acting on it are zero. This e-book covers translational and rotational equilibrium and practical worked examples for classroom and exam preparation.
Quick links: Edwin Ogie Library (home) • CBT Quiz
Forms of Equilibrium
Translational (Linear) Equilibrium
No net force — body at rest or moving with constant velocity (ΣFx = 0, ΣFy = 0).
Rotational Equilibrium
No net torque — sum of clockwise moments = sum of anticlockwise moments (ΣM = 0).
Resultant and Equilibrant Forces
The resultant is the vector sum of all forces. The equilibrant is equal in magnitude but opposite in direction to the resultant.
Parallelogram law — R² = P² + Q² + 2PQ cosθ
Example RE1: Two Forces (30 N & 40 N, θ = 60°)
Solution: R² = 30² + 40² + 2×30×40×cos60° = 900 + 1600 + 2400×0.5 = 3700 ⇒ R ≈ 60.8 N. Equilibrant = 60.8 N opposite direction.
Couples & Torque
A couple is two equal and opposite forces with parallel lines of action but not collinear — producing a pure rotation. Torque (moment) = Force × perpendicular distance (Nm)
Example CT1: Torque
50 N acting at 0.4 m produces torque = 50 × 0.4 = 20 Nm.
Equilibrium of Three Forces (Triangle of Forces)
If three forces acting at a point keep a body in equilibrium, they can be arranged to form a closed triangle.
Example TF1
Three forces are in equilibrium if the vector triangle closes (graphical or numerical check required).
Moment of a Force & Principle of Moments
Moment = Force × perpendicular distance
For equilibrium, sum of clockwise moments = sum of anticlockwise moments about any pivot.
Example MM1
A 40 N force at 0.5 m from pivot: Moment = 40 × 0.5 = 20 Nm.
Metre Rule Worked Example
Problem (i): A metre rule balances at 48 cm. A 60 g mass is at 6 cm and causes the balance point to shift to 30 cm. Find mass of the rule.
Moments about 30 cm: W×(48−30)=60×(30−6) ⇒ W×18 = 1440 ⇒ W = 80 g.
Problem (ii): Move the 60 g mass to 13 cm, find new balance point x (W=80 g).
60(x−13)=80(48−x) ⇒ 140x = 4620 ⇒ x ≈ 33 cm.
Additional Worked Examples
EQ1: Beam with two forces
A beam supported at center with 100 N and 150 N at 2 m and 3 m opposite sides — is it in equilibrium?
Clockwise 100×2 = 200 Nm, Anticlockwise 150×3 = 450 Nm ⇒ Not in equilibrium.
EQ2: Signboard with two cable tensions (vertical components)
T₁ sin40° + T₂ sin50° = 200 N. Additional data needed to find T₁ and T₂ individually.
Summary of Key Concepts
- Equilibrium: No net force and no net moment.
- Resultant & Equilibrant: Equilibrant balances the resultant.
- Moment & Torque: Force × perpendicular distance (Nm).
- Triangle of forces: Three forces in equilibrium form a closed triangle.
30-Question CBT Quiz (15 minutes)
Designed for JSS / early secondary students. Click Start Quiz to begin.
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