# Statics of rigid bodies

> Edexcel International A-Level Mathematics · IAL M2
> Source: https://www.owlsprep.com/study/edexcel-ial-math-m2-statics-of-rigid-bodies/

This guide covers Edexcel IAL M2 statics of rigid bodies, including moment calculations for non-parallel coplanar forces, equilibrium conditions, and ladder/rod problems with smooth/rough surfaces, aligned to the 2018 exam specification.

**Prerequisites:** [M1 parallel force moments](https://www.owlsprep.com/study/edexcel-ial-math-m1-moments-parallel-forces/); [M1 forces, friction and particle equilibrium](https://www.owlsprep.com/study/edexcel-ial-math-m1-forces-friction-equilibrium/)

## Learning objectives

- Calculate moments of coplanar non-parallel forces about a chosen pivot point
- Apply the two conditions for rigid body equilibrium to solve unknown force problems
- Solve ladder and rod problems involving smooth/rough contact using resolution, moments and $F \leq \mu R$
- Select optimal pivot points to simplify moment calculations in exam questions

## 1. Moment of a coplanar non-parallel force

**Moment of a force** — Turning effect of force $F$ about point $P$, equal to $F$ multiplied by the perpendicular distance from $P$ to the line of action of $F$. Units: N m.

*Notation:* $M$

*Example:* A 10 N force acting 2 m perpendicular to P has a moment of 20 N m clockwise.

For non-parallel forces acting at angles to rigid bodies, you can either calculate the perpendicular distance directly, or resolve the force into horizontal and vertical components and sum the moments of each component. This component method is almost always simpler for exam problems involving angled rods or ladders.

$$M = F \times d_\bot = F_x y + F_y x$$

Where $F_x, F_y$ are horizontal/vertical components of $F$, and $x,y$ are horizontal/vertical distances from the pivot to the point of force application. Always state a consistent sign convention for moments (e.g. anti-clockwise = positive) at the start of every problem.

**Worked example:** A uniform rod AB of length 4 m is pivoted at A. A 20 N force acts at B at 30° above the horizontal rod. Calculate the moment of the 20 N force about A.

1. Resolve the force into components perpendicular and parallel to the rod. The parallel component has zero moment as its line of action passes through A.
2. $$F_\bot = 20 \times \text{sin}(30^\text{o}) = 10 \text{ N (upwards)}$$
3. The perpendicular distance from A to the line of action of $F_\bot$ is equal to the length of the rod: 4 m.
4. $$M = 10 \times 4 = 40 \text{ N m (anti-clockwise)}$$
5. Verify with the component method: vertical force component 10 N acts 4 m horizontally from A, horizontal component acts 0 m vertically from A, so the same result is returned.

> **Exam tip:** Double-check that you only use the perpendicular component of force, or perpendicular distance, for moment calculations: using the full force magnitude or straight-line distance will give an incorrect result.

*Calculator:* allowed

## 2. Conditions for rigid body equilibrium

Unlike particle equilibrium, which only requires resultant force = 0, rigid bodies can rotate even if the net force is zero, so you must satisfy two equilibrium conditions for every problem:

1. Resultant force in all directions is zero: $\sum F_x = 0$, $\sum F_y = 0$
2. Resultant moment about any point in the plane is zero: $\sum M = 0$

**Coplanar force system** — A set of forces all acting in the same 2D plane, the only case covered in Edexcel IAL M2 statics of rigid bodies.

**Worked example:** A uniform rod AB of mass 5 kg and length 2 m is hinged at A and held horizontally by a vertical string attached at B. Find the tension in the string and the reaction force at the hinge.

1. Draw a force diagram: weight 5g N acts downwards at the midpoint of AB, tension $T$ N acts upwards at B, hinge reaction has horizontal $R_x$ and vertical $R_y$ components at A.
2. Apply horizontal force equilibrium: no horizontal forces act, so $R_x = 0$.
3. $$\text{Vertical equilibrium: } R_y + T - 5g = 0$$
4. Take moments about A to eliminate $R_x$ and $R_y$: clockwise moment from weight equals anti-clockwise moment from tension.
5. $$2T = 5g \times 1 \\ T = 2.5 \times 9.8 = 24.5 \text{ N}$$
6. Substitute $T$ back into vertical equilibrium: $R_y = 5g - 24.5 = 24.5$ N upwards. The hinge reaction is 24.5 N vertically upwards.

> **Exam tip:** Taking moments about a point where unknown forces act eliminates those forces from your equation, reducing the number of simultaneous equations you need to solve.

*Calculator:* allowed

## 3. Ladder and rod problems with smooth/rough contact

The most common exam problem type for this topic involves ladders or rods resting against walls and floors that may be smooth or rough. You will need to combine equilibrium conditions with the friction law $F \leq \mu R$ for rough surfaces.

> **Contact rule reminder**
>
> Smooth surfaces have zero friction, only normal reaction forces perpendicular to the surface. Rough surfaces have both normal reaction and friction parallel to the surface, opposing impending slip.

**Worked example:** A uniform ladder of mass 10 kg and length 6 m rests against a smooth vertical wall and rough horizontal ground. The ladder makes a 60° angle with the ground, and $\mu = 0.3$ between the ladder and ground. Determine if the ladder will slip.

1. Draw a force diagram: weight 10g N downwards at midpoint, normal reaction $R_w$ horizontal from smooth wall, normal reaction $R_g$ vertical upwards from ground, friction $F$ horizontal towards the wall at ground (opposing base slip away from wall).
2. $$\text{Horizontal equilibrium: } R_w - F = 0 \\ R_w = F$$
3. $$\text{Vertical equilibrium: } R_g - 10g = 0 \\ R_g = 10 \times 9.8 = 98 \text{ N}$$
4. Take moments about the ground contact point to eliminate $R_g$ and $F$: clockwise moment from weight equals anti-clockwise moment from $R_w$.
5. $$R_w \times 6 \text{sin}(60^\text{o}) = 10g \times 3 \text{cos}(60^\text{o}) \\ R_w = \frac{147}{3\sqrt{3}} \\ R_w \approx 28.3 \text{ N} = F$$
6. $$\text{Maximum friction: } F_{max} = \mu R_g = 0.3 \times 98 = 29.4 \text{ N}$$
7. Required $F = 28.3$ N < $F_{max}$, so the ladder will not slip.

> **Exam tip:** For ladder problems, always take moments about the ground contact point first: this eliminates both ground normal reaction and friction, leaving only one unknown to solve for directly.

*Calculator:* allowed

## 4. Exam problem solving strategy

1. Draw a clear, labelled force diagram marking all forces, their points of application and angles
2. State your moment sign convention (e.g. anti-clockwise = positive)
3. Write horizontal and vertical force equilibrium equations: $\sum F_x = 0$, $\sum F_y = 0$
4. Choose a pivot that eliminates the maximum number of unknown forces, write the $\sum M = 0$ equation
5. Solve the system of equations for unknowns
6. For friction problems, compare required $F$ to $\mu R$ to check for slip if requested

**Check your understanding**

Check your understanding of pivot choice

1. What is the optimal pivot to choose for a ladder problem if you need to calculate the normal reaction at the wall?

   *Why:* This eliminates the ground normal reaction and friction force from your moment equation, leaving only the wall normal reaction and weight as unknowns, so you can solve for the wall reaction directly.

> **Exam tip:** You will be awarded marks for drawing a clear force diagram in exams, even if you make an error in later calculations: always include this step.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using the full force magnitude instead of the perpendicular component for moment calculations
  - Why it fails: Only the component of force perpendicular to the line from the pivot to the point of application contributes to the turning effect
  - Correct: Either calculate the perpendicular distance from the pivot to the line of action of the force, or resolve the force into horizontal/vertical components and sum their individual moments
- **Wrong:** Omitting the moment equilibrium condition and only applying particle force equilibrium
  - Why it fails: Rigid bodies can rotate even if the net force is zero, so both equilibrium conditions must be satisfied to find all unknown forces
  - Correct: Always write two force equilibrium equations and one moment equilibrium equation for every rigid body statics problem
- **Wrong:** Adding a friction force at smooth contact points
  - Why it fails: Smooth surfaces have zero friction by definition, so only normal reaction forces perpendicular to the contact surface exist
  - Correct: Only include friction forces at explicitly marked rough contact points, acting parallel to the surface to oppose impending slip
- **Wrong:** Assuming friction force is always equal to $\mu R$ in all equilibrium problems
  - Why it fails: Friction only reaches its maximum value $\mu R$ when the body is on the point of slipping; for static equilibrium, friction is less than or equal to this value
  - Correct: Calculate required friction using equilibrium first, only compare to $\mu R$ if you are checking for slip or finding minimum $\mu$
- **Wrong:** Mixing up sign conventions for moments mid-calculation
  - Why it fails: Inconsistent sign conventions lead to incorrect moment sums and wrong unknown force values
  - Correct: State your sign convention clearly at the start of every problem, and apply it consistently to all moments in your calculation

## Cheatsheet

| Concept | Rule/Formula | Exam Use Case |
| --- | --- | --- |
| Moment of a force | $M = F \times d_\bot$, consistent sign convention | Calculate turning effect of any coplanar force |
| Rigid body equilibrium | $\sum F_x = 0$, $\sum F_y = 0$, $\sum M = 0$ | Solve for unknown reactions, tensions, forces |
| Smooth contact | Zero friction, only normal reaction $R$ perpendicular to surface | Problems with smooth walls/floors |
| Rough contact | $F \leq \mu R$, friction opposes impending slip | Slip checks, rough surface problems |
| Optimal pivot | Choose pivot with maximum unknown forces acting on it | Eliminate unknowns to simplify moment equations |

## What's next

Now that you have mastered statics of rigid bodies for Edexcel IAL M2, you are ready to progress to other core M2 topics including work-energy principles, projectile motion, and kinematics of variable acceleration. This topic is a foundational building block if you plan to study M3, where you will extend equilibrium concepts to 3D bodies and centre of mass calculations. To prepare for your exam, prioritise practicing full structured past paper questions for this topic to refine your force diagram drawing, moment calculation speed, and ability to select optimal pivot points. Ensure you consistently apply the correct sign convention for moments and friction rules to avoid losing easy marks.

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