How to Solve: Moments and Equilibrium

How to Solve: Moments and Equilibrium

GCSE & A-Level Maths

Introduction

"Master moments and equilibrium, and you’ll solve real-world problems—from balancing a seesaw to designing bridges—and ace 6-10 marks in your GCSE/A-Level Mechanics exam. This is one of the few topics where a single mistake can cost you the whole question, so let’s break it down step by step."

What You Need To Know First

Before diving in, make sure you understand: 1. Forces as vectors – Forces have size and direction (e.g., 5 N upwards vs. 5 N downwards). 2. Newton’s First Law – An object is in equilibrium if the resultant force is zero and the resultant moment is zero. 3. Basic trigonometry – Resolving forces into horizontal/vertical components (SOHCAHTOA).

Key Vocabulary

Formulas To Know

1. Moment of a Force
   Formula: Moment = Force × Perpendicular Distance
   Variables:
2. Moment (Nm) – Turning effect.
3. Force (N) – The push or pull.

4. Perpendicular Distance (m) – The shortest distance from the pivot to the force’s line of action.
   MEMORISE THIS – Not always given on the formula sheet.

5. Principle of Moments (Equilibrium Condition)
   Formula: Sum of CW Moments = Sum of ACW Moments
   MEMORISE THIS – The golden rule for equilibrium.

6. Resultant Force = 0 (for equilibrium)
   Formula: ΣF↑ = ΣF↓ (vertical) and ΣF→ = ΣF← (horizontal)
   Given on exam sheet (usually as "For equilibrium, resultant force = 0").

Step-by-Step Method

Step 1: Identify the Pivot

• Circle or label the pivot point on the diagram.
• Why? Moments are calculated around this point.

Step 2: List All Forces Acting on the Object

• Include weights, applied forces, and reaction forces at the pivot.
• Why? Missing a force = wrong answer.

Step 3: Resolve Forces into Perpendicular Distances

• For each force, find the perpendicular distance from the pivot to the force’s line of action.
• How?
• If the force is already perpendicular to the object (e.g., weight acting straight down), the distance is just the horizontal distance from the pivot.
• If the force is at an angle, use trigonometry (e.g., d × sinθ or d × cosθ) to find the perpendicular component.

Step 4: Calculate Each Moment

• Use Moment = Force × Perpendicular Distance.
• Label each moment as CW or ACW.
• Why? You’ll need to compare them in Step 5.

Step 5: Apply the Principle of Moments

• Set Sum of CW Moments = Sum of ACW Moments.
• Solve for the unknown (e.g., force, distance, or angle).

Step 6: Check for Equilibrium (Optional but Recommended)

• Ensure resultant force = 0 (ΣF↑ = ΣF↓ and ΣF→ = ΣF←).
• Why? Some questions ask for both conditions.

Worked Examples

Example 1 – Basic (GCSE Level)

Question: A 3 m uniform rod is pivoted at its centre. A 4 N weight is placed 1 m to the left of the pivot. What force must be applied 1.5 m to the right to balance the rod?

Step-by-Step Solution: 1. Pivot: Centre of the rod (1.5 m from each end). 2. Forces:
- 4 N (left, ACW).
- Unknown force F (right, CW).
- Weight of the rod (uniform = acts at centre, so no moment). 3. Perpendicular distances:
- 4 N: 1 m from pivot.
- F: 1.5 m from pivot. 4. Moments:
- ACW: 4 N × 1 m = 4 Nm.
- CW: F × 1.5 m. 5. Principle of Moments:
4 Nm = F × 1.5 m
F = 4 / 1.5 = 2.67 N (2 d.p.) 6. Check equilibrium:
- Vertical forces: 4 N (down) + F (down) = 6.67 N.
- Reaction at pivot = 6.67 N (up) → ΣF↑ = ΣF↓.

What we did and why: We used the principle of moments to balance the turning effects. The rod’s weight didn’t matter because it acted at the pivot (zero moment).

Example 2 – Medium (A-Level Level)

Question: A 5 m ladder leans against a smooth wall at 60° to the ground. The ladder has a mass of 20 kg. A 70 kg painter stands 3 m up the ladder. Find the reaction force at the wall.

Step-by-Step Solution: 1. Pivot: Base of the ladder (ground contact). 2. Forces:
- Weight of ladder (20 kg × 9.8 = 196 N) – acts at 2.5 m (centre).
- Weight of painter (70 kg × 9.8 = 686 N) – acts at 3 m.
- Reaction at wall (R) – horizontal (smooth wall = no friction).
- Reaction at ground (N and F) – vertical and horizontal. 3. Perpendicular distances:
- Ladder weight: 2.5 m × cos(60°) = 1.25 m from pivot.
- Painter weight: 3 m × cos(60°) = 1.5 m from pivot.
- Wall reaction: 5 m × sin(60°) = 4.33 m from pivot. 4. Moments (ACW = CW):
- ACW: (196 N × 1.25 m) + (686 N × 1.5 m) = 245 + 1029 = 1274 Nm.
- CW: R × 4.33 m. 5. Solve for R:
1274 = R × 4.33
R = 1274 / 4.33 ≈ 294 N 6. Check equilibrium (horizontal):
- R (wall) = 294 N → F (ground friction) = 294 N (opposite direction).

What we did and why: We resolved forces at an angle using trigonometry and took moments around the base to avoid dealing with the ground reaction forces. The wall’s reaction was horizontal because it was smooth (no friction).

Example 3 – Exam-Style (Time-Pressure)

Question: A 4 m beam is pivoted 1 m from its left end. A 10 N force acts downward at the left end, and a 15 N force acts downward at the right end. Where must a 20 N force be placed to balance the beam?

Step-by-Step Solution: 1. Pivot: 1 m from left end (label it). 2. Forces:
- 10 N (left, ACW).
- 15 N (right, CW).
- 20 N (unknown position, x m from pivot, ACW or CW?). 3. Perpendicular distances:
- 10 N: 1 m from pivot (ACW).
- 15 N: 3 m from pivot (CW) (4 m total – 1 m pivot = 3 m).
- 20 N: x m from pivot. 4. Assume 20 N is ACW (if wrong, x will be negative):
- ACW: (10 N × 1 m) + (20 N × x).
- CW: 15 N × 3 m = 45 Nm. 5. Principle of Moments:
10 + 20x = 45
20x = 35
x = 1.75 m 6. Interpretation:
- x = 1.75 m from pivot (toward the right end).
- Total distance from left end = 1 m (pivot) + 1.75 m = 2.75 m from left end.

What we did and why: We assumed the 20 N force acted ACW first. If x had been negative, we’d flip the direction. The answer is the distance from the left end, not the pivot—read the question carefully!

Common Mistakes

Exam Traps

1-Minute Recap

"Here’s the night-before cheat sheet for moments and equilibrium: 1. Pivot first – Circle it on the diagram. 2. List all forces – Weights, applied forces, reactions. 3. Perpendicular distance – Use trig if the force is at an angle. 4. Moment = Force × Distance – Label CW or ACW. 5. Sum CW = Sum ACW – Solve for the unknown. 6. Check equilibrium – ΣF↑ = ΣF↓ and ΣF→ = ΣF←. Common traps? Angled forces, hidden weights, and mixing up distances. Double-check your units (Nm, N, m) and always draw a diagram. You’ve got this!