Equilibrium of Forces part 2
Edwin Ogie Library
Equilibrium of Forces E‑Note
Objectives:
• Apply the conditions for equilibrium of coplanar forces.
• Use triangle and polygon laws of forces to solve equilibrium problems.
• Apply Lami’s theorem to solve problems.
• Analyze the principle of moment of a force and determine moments (including couples/torque).
• Resolve forces into two perpendicular directions and determine resultant and equilibrant.
• Differentiate between stable, unstable, and neutral equilibrium.
Page 1: Introduction to Equilibrium of Forces
- Equilibrium of particles – including coplanar forces, triangles and polygons of forces, and Lami’s theorem.
- Principles of moments – moment of a force, moment of a couple (torque), and their applications.
- Equilibrium of rigid bodies – resolving and composing forces, and finding the resultant and equilibrant.
- Centre of gravity and stability – distinguishing between stable, unstable, and neutral equilibrium.
Page 2: Equilibrium of Coplanar Forces
A particle is in equilibrium if the vector sum of all forces acting on it is zero. Coplanar forces are forces that lie in the same geometric plane. This means all the forces act in a two-dimensional space, such as on a flat surface. These forces can be: - Concurrent: All forces meet at a single point. - Parallel: Forces run in the same direction or opposite directions but never meet. - Non-concurrent and non-parallel: Forces lie in the same plane but neither meet nor run parallel. For a body to remain in equilibrium under the action of coplanar forces, two key conditions must be satisfied: 1. Resultant Force = 0 The vector sum of all forces acting on the body must be zero. 2. Resultant Moment = 0 The sum of all moments about any point must also be zero. For coplanar forces:
- ΣFx = 0
- ΣFy = 0
This means the forces balance in both the horizontal and vertical directions.
Example CF1: Equilibrium of Three Forces
Solution:
Given three forces acting on a particle, if F₁, F₂, and F₃ are in equilibrium then F₁ + F₂ + F₃ = 0. As illustrated in the diagram above.
Page 3: Triangle and Polygon of Forces
When several forces act concurrently on a particle, they can be arranged head-to-tail to form a closed polygon if the particle is in equilibrium. In a triangle of forces, the magnitudes of the forces satisfy the triangle rule.
Example TP1: Triangle of Forces
Solution:
Suppose three forces of magnitudes 10 N, 15 N, and an unknown force F act at a point and form a triangle. If the triangle closes, then F can be found by applying the sine rule to the triangle.
Example TP2: Polygon of Forces
Solution:
For a polygon of forces, if forces F₁, F₂, F₃, …, Fn form a closed polygon then their vector sum is zero. Use geometry to solve for unknown forces.
Example TP3: Constructing a Force Polygon
Solution:
Draw each force vector head-to-tail on graph paper. The closing side represents the equilibrant force. Measure its length and direction to determine its magnitude and bearing.
Page 4: Lami’s Theorem
Lami’s theorem applies to a particle in equilibrium under three coplanar, concurrent forces. It states that:Lami’s Theorem states that when three coplanar, concurrent, non-collinear forces \(Fa\), \(Fb\), and \(F_c\) act on a particle in equilibrium, each force is proportional to the sine of the angle between the other two forces.
F1/sin α = F2/sin β = F3/sin γ
where α, β, and γ are the angles between each force and the line of action of the resultant of the other two.
Example LT1: Applying Lami’s Theorem
Solution:
Given forces F1 = 20 N, F2 = 30 N, and the angles between them are provided, use Lami’s theorem to calculate F3.
For example, if sin α = 0.5 and sin β = 0.6, then F3 can be determined by the equality F1/sin α = F3/sin γ.
Example LT2: Determining an Unknown Force
Solution:
If F1 = 25 N, F2 = 35 N, and the angles opposite to them are 40° and 50° respectively, then using Lami’s theorem the unknown force F3 can be calculated.
Example LT3: Verifying Equilibrium
Solution:
Check that F1/sin α = F2/sin β = F3/sin γ holds for the given forces to confirm equilibrium.
Page 5: Principles of Moments – Moment of a Force
The moment of a force about a point (or axis) is the turning effect produced by the force. It is defined as:
Moment (M) = Force (F) × Perpendicular distance (d)
For a force to produce rotation about a pivot, only its component perpendicular to the line joining the point and the line of action matters.
Example PM1: Calculating a Simple Moment
Solution:
If F = 50 N and the perpendicular distance d = 0.8 m, then Moment = 50 × 0.8 = 40 N·m.
Example PM2: Moment about a Pivot
Solution:
For a force of 30 N acting at 60° to a lever of length 1.5 m, first find the perpendicular component: d = 1.5 × sin 60° ≈ 1.5 × 0.866 = 1.299 m. Moment ≈ 30 × 1.299 ≈ 38.97 N·m.
Example PM3: Moment Calculation with Diagram
Solution:
Given a diagram with a force F = 40 N and a measured perpendicular distance of 1.2 m, Moment = 40 × 1.2 = 48 N·m.
Page 6: Moment of a Couple and Applications
A couple consists of two equal and opposite forces whose lines of action do not coincide. This setup doesn’t cause linear motion—but it does produce rotation. The moment (or torque) of a couple is given by:
Moment (M) = Force (F) × Distance (d)
Note that the moment of a couple is independent of the pivot point.
Example MC1: Basic Couple Moment
Solution:
Two forces of 20 N separated by 0.5 m produce a moment = 20 × 0.5 = 10 N·m.
Example MC2: Seesaw Example
Solution:
On a seesaw, if a child of 300 N sits 1.2 m from the pivot and another of 250 N sits 1.5 m on the other side, the moments are 300×1.2 and 250×1.5. Compare to check for balance.
Example MC3: Torque in a Rotating System
Solution:
If a wrench applies a force of 50 N at a perpendicular distance of 0.3 m from the bolt, torque = 50 × 0.3 = 15 N·m.
Page 7: Equilibrium of Rigid Bodies – Resolution and Composition
For a rigid body to be in equilibrium, both the resultant force and the resultant moment must be zero. Forces can be resolved into two perpendicular directions (usually horizontal and vertical):
Fx = F cos θ, Fy = F sin θ
Example ER1: Resolving a Force
Solution:
A force of 100 N at 30° above the horizontal has Fx = 100 cos 30° ≈ 86.6 N and Fy = 100 sin 30° = 50 N.
Example ER2: Composition of Two Forces
Solution:
If forces of 40 N (east) and 30 N (north) act on a body, the resultant = √(40²+30²)= 50 N at an angle tan−1(30/40) ≈ 36.87° north of east.
Example ER3: Using Components for Equilibrium
Solution:
In a system with forces acting in perpendicular directions, set ΣFx = 0 and ΣFy = 0 to solve for unknown forces.
Page 8: Resultant and Equilibrant
The resultant of a set of forces is their vector sum, and the equilibrant is a force equal in magnitude but opposite in direction to the resultant. The resultant is the single force that has the same effect as two or more forces acting together. It’s found by vector addition of all individual forces. The equilibrant is the force that exactly balances the resultant. It’s equal in magnitude but opposite in direction
Example RE1: Finding the Resultant of Two Forces
Solution:
For forces of 30 N east and 40 N north, the resultant R = √(30² + 40²) = 50 N at an angle = tan−1(40/30) ≈ 53.13° north of east.
Example RE2: Determining the Equilibrant
Solution:
The equilibrant is equal in magnitude but opposite in direction to the resultant. In Example RE1, the equilibrant is 50 N at 180° + 53.13° = 233.13° (measured from east).
Example RE3: Using Components to Find the Resultant
Solution:
If F1 = 25 N at 20° above horizontal and F2 = 30 N at 10° below horizontal, resolve each into components, sum them, and then find the magnitude and direction of the resultant.
Page 9: Centre of Gravity and Stability
The centre of gravity of a body is the point at which its weight is considered to be concentrated. A body is said to be:
- Stable: When displaced, it returns to its original position.
- Unstable: When displaced, it moves further away from equilibrium.
- Neutral: When displaced, it remains in the new position.
Example CG1: Determining Centre of Gravity
Solution:
In a uniform rod, the centre of gravity is at its midpoint. For a 4 m rod, the centre is at 2 m.
Example CG2: Stability of a Freestanding Object
Solution:
An object with a low centre of gravity relative to its base is stable. For instance, a wide, low table is more stable than a tall, narrow one.
Example CG3: Neutral Equilibrium
Solution:
A ball on a flat surface can be moved to a new position and remain there; this is an example of neutral equilibrium.
Page 10: Summary of Key Formulas and Concepts
- Equilibrium of forces: ΣF = 0 (both x and y components).
- Cosine rule for force polygons.
- Lami’s Theorem: F1/sin α = F2/sin β = F3/sin γ.
- Moment of a force: M = F × d.
- Moment of a couple: M = F × d (independent of pivot).
- Resolution of forces: Fx = F cos θ, Fy = F sin θ.
- Resultant and Equilibrant: Vector sum and its opposite.
- Centre of Gravity: Point where weight is concentrated; Stability: Determined by relative position of CG and base.
Page 11: Extended Discussion and Applications
The principles of equilibrium are applied in structural engineering, bridge design, and machinery. By ensuring that forces and moments balance, safe and efficient designs are achieved. Lami’s theorem is particularly useful in solving problems where three forces act on a point, while the moment principles underpin the analysis of levers and torque.
Page 12: Additional Tips and Insights
Always draw free body diagrams to visualize forces. Check that the sum of horizontal and vertical components equals zero for equilibrium. Use the sine and cosine rules to resolve forces in force polygons and verify results with Lami’s theorem when applicable.
Page 13: Final Summary
- A particle or rigid body is in equilibrium if the sum of forces and moments is zero.
- For coplanar forces, both ΣFx = 0 and ΣFy = 0 must hold.
- Lami’s theorem provides a relation between three concurrent forces in equilibrium.
- The moment of a force is the product of the force and its perpendicular distance from the pivot.
- Resultant and equilibrant are key to understanding how forces balance.
- Stability depends on the position of the centre of gravity relative to the base of support.
Page 14: Quiz Introduction
Test your understanding of the equilibrium of forces by attempting the following quiz. You will have 15 minutes to answer 30 questions.
30 CBT JAMB Quiz on Equilibrium of Forces
Click the "Start Quiz" button to begin. You will have 15 minutes to answer 30 questions.
RSS Feed
Subscribe to our RSS feed for the latest updates on our Equilibrium of Forces enote lessons and quizzes:
Edwin Ogie Library - Equilibrium of Forces Enote http://www.edwinogie-library.comLatest updates on equilibrium of forces, moments, and stability. New Lesson on Equilibrium of Forces http://www.edwinogie-library.com/equilibrium-of-forcesExplore our comprehensive enote covering coplanar forces, Lami’s theorem, moments, and stability.
Comments