# Statics of a Particle (Edexcel IAL M1)

> Edexcel International A-Level Mathematics · IAL M1
> Source: https://www.owlsprep.com/study/edexcel-ial-math-m1-statics-of-a-particle/

This guide covers core statics of a particle content for Edexcel IAL M1, including force resolution, particle equilibrium conditions, and friction rules for static problems, with exam-aligned worked examples and mistake avoidance tips.

**Prerequisites:** [Forces as vectors and vector addition (Edexcel IAL M1 T03)](https://www.owlsprep.com/study/edexcel-ial-math-m1-forces-as-vectors/); Basic right-angled triangle trigonometry (SOHCAHTOA)

## Learning objectives

- Resolve coplanar forces into perpendicular components accurately
- Apply particle equilibrium conditions to form valid force equations
- Distinguish between static and limiting friction using the F ≤ μR rule
- Solve exam-standard statics problems involving weight, normal reaction, tension, thrust and friction

## 1. Resolution of Coplanar Forces

**Coplanar forces** — Forces that all act in the same 2D plane, the only case covered in M1 particle statics

To resolve a force $F$ acting at angle $\theta$ to a chosen axis, split it into two perpendicular components: the component *along* the axis is $F\cos\theta$, and the component *perpendicular* to the axis is $F\sin\theta$. For slope problems, choose axes parallel and perpendicular to the incline instead of horizontal/vertical to simplify calculations. Always draw a labelled force diagram first.

**Worked example:** A force of 20 N acts on a particle at 30° above the horizontal. Find the horizontal and vertical components of the force, giving answers to 3 significant figures.

1. Step 1: Identify force $F = 20$ N, angle from horizontal $\theta = 30^\circ$
2. $$F_x = 20\cos30^\circ = 10\sqrt{3} ≈ 17.3 \text{ N (horizontal component, rightwards)}$$
3. $$F_y = 20\sin30^\circ = 10 \text{ N (vertical component, upwards)}$$

> **tip**
>
> Double-check your angle reference: if $\theta$ is between the force and the axis you are resolving along, use $\cos\theta$ for that component, $\sin\theta$ for the perpendicular component.

*Calculator:* allowed

## 2. Equilibrium of a Particle

**Particle equilibrium** — A particle is in equilibrium if the resultant of all coplanar forces acting on it is zero, meaning the sum of forces along each perpendicular axis equals 0.

Common forces you will encounter in equilibrium problems: weight ($W = mg$, acts vertically downwards), normal reaction ($R$, acts perpendicular to the contact surface, pushing on the particle), tension (pull force in a string/rope, acts along the string away from the particle), thrust (push force in a rigid rod, acts along the rod towards the particle).

**Worked example:** A 2 kg particle is suspended at rest by two identical vertical strings that share the weight equally. Find the tension in each string.

1. Step 1: Calculate weight: $W = 2 \times 9.8 = 19.6$ N downwards. Draw force diagram with weight and two equal upward tensions $T_1, T_2$.
2. Step 2: Resolve vertically upwards: sum of upward forces = sum of downward forces, so $T_1 + T_2 = 19.6$
3. Step 3: Tensions are equal, so $T_1 = T_2 = 19.6 / 2 = 9.8$ N each.

> **Exam tip**
>
> Always state your equilibrium equations clearly, labelling the direction you are resolving (e.g., "Resolving parallel to the slope, upwards positive:") to earn method marks even if your final answer is incorrect.

*Calculator:* allowed

## 3. Static Friction and Limiting Equilibrium

**Limiting friction** — The maximum static frictional force between two stationary surfaces, given by $F_{max} = \mu R$. For particles in equilibrium that are not on the point of moving, $F \leq \mu R$.

Friction always acts in the direction opposite to the direction the particle would move if there was no friction. Only use $F = \mu R$ if the question states the particle is "on the point of sliding", "in limiting equilibrium" or "just about to move". For all other static cases, frictional force equals the magnitude of the force it is opposing, which is less than $\mu R$.

**Worked example:** A 3 kg block rests on a rough horizontal surface with coefficient of friction $\mu = 0.2$. A horizontal force of 5 N is applied to the block, which remains stationary. Find the frictional force acting on the block, and verify it is below the limiting friction value.

1. Step 1: Resolve vertically: normal reaction $R = W = 3 \times 9.8 = 29.4$ N
2. $$F_{max} = \mu R = 0.2 \times 29.4 = 5.88 \text{ N (limiting friction)}$$
3. Step 2: Resolve horizontally: 5 N applied force right, so friction $F = 5$ N left to balance the applied force.
4. Step 3: Verify $5 < 5.88$, so the block remains at rest as stated.

> **warning**
>
> Never automatically use $F = \mu R$ for static problems unless the question explicitly states the particle is at limiting equilibrium. This is one of the most common errors in M1 statics questions.

*Calculator:* allowed

## 4. Combined Statics Exam Problems

Most exam statics questions combine force resolution, equilibrium, and friction, often involving particles on inclined planes or connected by strings. Always start with a clear, fully labelled force diagram before starting calculations.

**Worked example:** A 5 kg particle is on the point of sliding UP a rough plane inclined at 20° to the horizontal, pulled by a string parallel to the plane up the slope (limiting equilibrium, maximum tension). The coefficient of friction between the particle and plane is 0.3. Find the tension in the string, giving your answer to 3 significant figures.

1. Step 1: Draw force diagram: weight $W = 5 \times 9.8 = 49$ N downwards, normal reaction $R$ perpendicular to the plane upwards, tension $T$ up the plane, friction $F$ down the plane (the particle is on the point of sliding UP the plane, so friction opposes that upward motion).
2. $$\text{Resolve perpendicular to plane: } R = W\cos20^\circ = 49\cos20^\circ ≈ 46.04 \text{ N}$$
3. $$\text{Limiting equilibrium, so } F = \mu R = 0.3 \times 46.04 ≈ 13.81 \text{ N}$$
4. $$\text{Resolve parallel to plane, upwards positive: } T = W\sin20^\circ + F ≈ 49\sin20^\circ + 13.81 ≈ 30.6 \text{ N}$$

> **Exam tip**
>
> Give all final answers to 2 or 3 significant figures, as per Edexcel marking guidance. Always use $g = 9.8$ m s⁻², do not use 9.81 or 10 unless explicitly told to in the question.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using $F = \mu R$ for all static friction problems
  - Why it fails: $F = \mu R$ only applies when the particle is in limiting equilibrium (on the point of moving). For stationary particles not at limiting friction, $F$ equals the opposing force, which is less than $\mu R$.
  - Correct: Only use $F = \mu R$ if the question explicitly states the particle is just about to move, in limiting equilibrium, or on the point of sliding.
- **Wrong:** Swapping $\sin$ and $\cos$ when resolving forces at an angle to the vertical
  - Why it fails: The component of a force along the axis it makes angle $\theta$ with is always $F\cos\theta$, regardless of whether the axis is horizontal or vertical.
  - Correct: Check the angle reference: if $\theta$ is between the force and the axis you are resolving along, use $\cos\theta$ for that component, $\sin\theta$ for the perpendicular component.
- **Wrong:** Assuming normal reaction $R$ is always equal to weight $mg$
  - Why it fails: On inclined planes, or when applied forces have vertical components, normal reaction is not equal to weight.
  - Correct: Always derive $R$ by resolving forces perpendicular to the contact surface, rather than assuming $R = mg$.
- **Wrong:** Drawing friction in the wrong direction
  - Why it fails: Friction opposes the direction the particle would move if there was no friction, not the direction of any applied force.
  - Correct: Before labelling friction, ask: "Which way would the particle move if friction was removed?" Friction acts opposite to that direction.
- **Wrong:** Using $g = 9.81$ or $10$ m s⁻² instead of $9.8$ m s⁻²
  - Why it fails: Edexcel IAL M1 explicitly requires use of $g = 9.8$ m s⁻², so using a different value will lead to loss of accuracy marks.
  - Correct: Always substitute $g = 9.8$ m s⁻² in all calculations unless the question specifies a different value.

## Cheatsheet

| Concept | Rule | Exam Reminder |
| --- | --- | --- |
| Force resolution | Component along axis = $F\cos\theta$, perpendicular = $F\sin\theta$ (θ = angle between force and axis) | Choose axes aligned to the slope for incline problems to simplify equations |
| Particle equilibrium | Sum of forces along each perpendicular axis = 0 | Label the direction you are resolving clearly to get method marks |
| Static friction | $F ≤ \mu R$ for stationary particles; $F = \mu R$ only for limiting equilibrium | Never use $F = \mu R$ unless the question states the particle is on the point of moving |
| Common forces | $W = mg$ (down), $R$ (perpendicular to surface), tension (pull along string), thrust (push along rod) | Draw a full labelled force diagram before starting any calculation |

## What's next

Now that you have mastered statics of a particle, you are ready to move on to dynamics of a particle, the next core Edexcel IAL M1 topic. In dynamics, you will use the same force resolution skills you practiced here, alongside Newton’s laws of motion, to solve problems involving moving particles, where $F = \mu R$ applies for all cases of motion between rough surfaces. You will build on these equilibrium skills later in M2 when you study rigid body statics, which covers extended objects like rods and ladders, but remember that all M1 statics questions are exclusively for point particles, with no moments required. Regular practice of past paper statics questions will help you avoid common mistakes and speed up your problem-solving for the exam.

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