Statics of rigid bodies

Q: What pivot point should I choose for rigid body statics problems?

Select a pivot where the maximum number of unknown forces act: this eliminates those forces from your moment equation, as their perpendicular distance from the pivot is zero, so their moment is zero. Common choices include hinges, ladder-ground contact points, or support reaction points.

Q: What are the two conditions for rigid body equilibrium?

1. The resultant of all coplanar forces is zero (, ). 2. The algebraic sum of moments of all forces about any point in the plane is zero (no net turning effect).

Q: When is friction force equal to ?

Friction only reaches its maximum value when the body is on the point of slipping. For static equilibrium where slip is not impending, friction force is less than or equal to , calculated using equilibrium conditions.

Edexcel International A-Level Mathematics· 2018 Issue 3 M2 §5.1-5.2· 25 min read

1. 1. Moment of a coplanar non-parallel force★★☆☆☆⏱ 6 min

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📘 Definition

Moment of a force

$M$

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.

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 $F_{⊥}$ 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.

2. 2. Conditions for rigid body equilibrium★★★☆☆⏱ 7 min

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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:

Resultant force in all directions is zero: $\sum F_{x} = 0$ , $\sum F_{y} = 0$
Resultant moment about any point in the plane is zero: $\sum M = 0$

📘 Definition

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: $R_{y} = 5 g - 24.5 = 24.5$ 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.

3. 3. Ladder and rod problems with smooth/rough contact★★★★☆⏱ 8 min

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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 $F \leq μ R$ for rough surfaces.

📐 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 $μ = 0.3$ 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 < $F_{ma x}$ , 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.

4. 4. Exam problem solving strategy★★★☆☆⏱ 4 min

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

✓ Quick check

Check your understanding of pivot choice

What is the optimal pivot to choose for a ladder problem if you need to calculate the normal reaction at the wall?
Reveal answer
The contact point between the ladder and the ground —
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.

5. Common Pitfalls

Wrong move:

Using the full force magnitude instead of the perpendicular component for moment calculations

Why:

Only the component of force perpendicular to the line from the pivot to the point of application contributes to the turning effect

Correct move:

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 move:

Omitting the moment equilibrium condition and only applying particle force equilibrium

Why:

Rigid bodies can rotate even if the net force is zero, so both equilibrium conditions must be satisfied to find all unknown forces

Correct move:

Always write two force equilibrium equations and one moment equilibrium equation for every rigid body statics problem

Wrong move:

Adding a friction force at smooth contact points

Why:

Smooth surfaces have zero friction by definition, so only normal reaction forces perpendicular to the contact surface exist

Correct move:

Only include friction forces at explicitly marked rough contact points, acting parallel to the surface to oppose impending slip

Wrong move:

Assuming friction force is always equal to $μ R$ in all equilibrium problems

Why:

Friction only reaches its maximum value $μ R$ when the body is on the point of slipping; for static equilibrium, friction is less than or equal to this value

Correct move:

Calculate required friction using equilibrium first, only compare to $μ R$ if you are checking for slip or finding minimum $μ$

Wrong move:

Mixing up sign conventions for moments mid-calculation

Why:

Inconsistent sign conventions lead to incorrect moment sums and wrong unknown force values

Correct move:

State your sign convention clearly at the start of every problem, and apply it consistently to all moments in your calculation

6. Quick Reference Cheatsheet

Concept	Rule/Formula	Exam Use Case
Moment of a force	$M = F \times d_{⊥}$ , 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	$F \leq μ R$ , 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

7. Frequently Asked

What pivot point should I choose for rigid body statics problems?

Select a pivot where the maximum number of unknown forces act: this eliminates those forces from your moment equation, as their perpendicular distance from the pivot is zero, so their moment is zero. Common choices include hinges, ladder-ground contact points, or support reaction points.

What are the two conditions for rigid body equilibrium?

The resultant of all coplanar forces is zero ( $\sum F_{x} = 0$ , $\sum F_{y} = 0$ ). 2. The algebraic sum of moments of all forces about any point in the plane is zero (no net turning effect).

When is friction force equal to $μ R$ ?

Friction only reaches its maximum value $μ R$ when the body is on the point of slipping. For static equilibrium where slip is not impending, friction force is less than or equal to $μ R$ , calculated using equilibrium conditions.

Going deeper

official_documentEdexcel IAL Maths 2018 SpecificationOfficial syllabus for M2 statics content
study_guideM1 Parallel Force Moments Recap

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.