What is the definition of statics of a particle in Mechanics 1?

In Mechanics 1, statics of a particle refers to the study of particles at rest or in equilibrium under the action of forces. This involves analyzing the conditions under which a particle remains stationary or moves with constant velocity, ensuring that the net force acting on the particle is zero. This concept is crucial in understanding how forces interact in systems studied in Edexcel International A Levels Mathematics.

How do you determine if a particle is in equilibrium?

A particle is in equilibrium when the vector sum of all forces acting on it is zero. This means that the forces balance each other out, resulting in no net force and, consequently, no acceleration. In mathematical terms, this is expressed as ΣF = 0, where ΣF represents the sum of all forces. This principle is fundamental in the statics of a particle, a key topic in Mechanics 1 for Edexcel International A Levels.

What role do free-body diagrams play in solving statics problems?

Free-body diagrams are essential tools in solving statics problems as they visually represent all the forces acting on a particle. By isolating the particle and illustrating forces such as tension, weight, and normal force, students can better understand the interactions and apply equilibrium conditions effectively. This method is widely used in Mechanics 1 to simplify complex problems and is a critical skill for Edexcel International A Levels Mathematics students.

How does the concept of equilibrium apply to inclined planes in statics?

In the context of statics, when a particle is on an inclined plane, equilibrium is achieved when the component of gravitational force parallel to the plane is balanced by other forces, such as friction or tension. This requires resolving forces into components parallel and perpendicular to the plane and ensuring that the net force in each direction is zero. Understanding this application is vital for solving inclined plane problems in Mechanics 1 for Edexcel International A Levels.

What are common mistakes to avoid when analyzing statics of a particle problems?

Common mistakes in statics of a particle problems include neglecting to consider all forces acting on the particle, incorrectly resolving forces into components, and failing to apply the equilibrium condition correctly. Students should ensure that they account for all forces, use accurate trigonometric functions for resolving forces, and verify that the sum of forces equals zero. Mastery of these aspects is crucial for success in Mechanics 1 within the Edexcel International A Levels Mathematics curriculum.

Why is "Drawing friction in the SAME direction as the tendency of motion" a common mistake in Statics of a particle?

Why it happens: Confusing 'tendency of motion' with the direction friction acts. How to avoid it: Friction OPPOSES the tendency of motion. If a force pushes the particle UP a slope, friction acts DOWN the slope. If gravity pulls the particle DOWN, friction acts UP.

Why is "Using the angle BETWEEN a force and one of the others, instead of the angle OPPOSITE the force in Lami's theorem" a common mistake in Statics of a particle?

Why it happens: Lami's theorem looks deceptively simple. How to avoid it: In Lami's P_i / \alpha_i, the angle \alpha_i is the angle between the OTHER two forces (the angle 'opposite' force P_i at the point of concurrency). Source: WME01/01 examiner reports

Why is "Solving for only one value of $P$ on a rough incline problem" a common mistake in Statics of a particle?

Why it happens: Not realising friction can act either way. How to avoid it: On a rough incline with applied force, work out BOTH cases: friction up the slope (about to slide down) and friction down the slope (about to slide up). The two values bracket the equilibrium range.

Why is "Using $R = mg\cos\theta$ when an applied force has a perpendicular component" a common mistake in Statics of a particle?

Why it happens: Memorising the formula without re-deriving. How to avoid it: On a slope with applied force P at some angle, always RE-DERIVE R by resolving perpendicular to the slope. R may include a P term depending on whether P pushes into or pulls away from the slope.

Why is "Using $\sin$ instead of $\cos$ (or vice versa) when resolving" a common mistake in Statics of a particle?

Why it happens: Forgetting the geometric convention. How to avoid it: Mark the angle on the diagram. The component **adjacent** to the angle is ; the component **opposite** to the angle is .

Mechanics 1Edexcel International A Levels Mathematics (XMA01-YMA01)

Statics of a particle

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next.

Previous Next

Learn this Topic

Start with the slides for the quick version, then go deeper with the full study notes.

Short Study Notes in the form of Slides

Read the notes first. If the method in a worked example clicks, you're ready for the questions.

Page 1 / 0

Detailed Study Notes

Full prose, callouts and a recap — built for A* mastery, not just a quick scan.

Take these study notes with you

Download a branded PDF — full prose, callouts, recap and memorise list for Statics of a particle, ready to print or save offline.

Statics of a Particle Study Notes — Edexcel IAL Mathematics M1 (WME01/01, 2018 spec — 2026 onwards)

A particle is in equilibrium when the resultant force is zero. Resolve coplanar forces into two perpendicular directions and set each algebraic sum to zero. For exactly three forces, the triangle-of-forces / Lami's theorem gives a slick alternative. Includes equilibrium on a rough inclined plane.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

M1 7.1 — Recognise the conditions for equilibrium of a particle under coplanar forces.
M1 7.2 — Resolve coplanar forces and set up two equations ( $\Sigma F_x = 0$ , $\Sigma F_y = 0$ ) to find unknown forces.
M1 7.3 — Apply the triangle-of-forces method and Lami's theorem for three concurrent forces in equilibrium.

Conditions for equilibrium

Two perpendicular sums of zero. State the convention.

A particle is in equilibrium when the resultant force acting on it is zero:

$\Sigma \mathbf{F} = \mathbf{0}.$

Equivalently, in any pair of perpendicular directions:

$\Sigma F_x = 0, \qquad \Sigma F_y = 0.$

This is enough for a PARTICLE. For a rigid body, you also need $\Sigma M = 0$ about any point (covered in the Moments topic).

Workflow:

Draw a force diagram with EVERY force acting on the particle, at the correct angle.
Choose axes. Often horizontal/vertical; on an inclined plane, along/perpendicular to the slope.
Resolve each force into the two axes.
Write $\Sigma F_x = 0$ and $\Sigma F_y = 0$ .
Solve the two simultaneous equations for the unknowns.

Worked example. Particle of mass $4$ kg hung by two strings from a ceiling; one at $30^{\circ}$ from vertical, the other at $60^{\circ}$ from vertical. Find tensions $T_A$ and $T_B$ .

Horizontal: $T_A \sin 30^{\circ} = T_B \sin 60^{\circ} \Rightarrow T_A = \sqrt{3} T_B$ .
Vertical: $T_A \cos 30^{\circ} + T_B \cos 60^{\circ} = 4g$ .
Substituting: $\tfrac{3}{2} T_B + \tfrac{1}{2} T_B = 39.2 \Rightarrow T_B = 19.6$ N, $T_A \approx 33.95$ N.

$\Sigma F_x = 0$ AND $\Sigma F_y = 0$ .
Two equations $\Rightarrow$ at most two unknowns.
Pick axes to simplify the resolved forces.
On an incline: axes along/perpendicular to slope.

Triangle of forces (3 forces in equilibrium)

Three forces $\Leftrightarrow$ closed triangle. Sine rule on the triangle.

Result. Three coplanar concurrent forces are in equilibrium if and only if they form a closed triangle when drawn head-to-tail.

This is geometrically obvious: equilibrium means the vector sum is zero; vector addition is geometrically tip-to-tail, and a zero sum returns to the start.

Sine rule on the triangle:

$\dfrac{P_1}{\sin\theta_1} = \dfrac{P_2}{\sin\theta_2} = \dfrac{P_3}{\sin\theta_3},$

where $\theta_i$ is the angle of the triangle OPPOSITE side $P_i$ .

Mapping angles. The angle 'between $P_2$ and $P_3$ ' at the particle EXTERNALLY is $180^{\circ} - \theta_1$ (the triangle's INTERNAL angle opposite $P_1$ ). When the question gives angles between forces (at the particle), you typically subtract from $180^{\circ}$ to get the triangle's internal angles.

Worked example. Three forces in equilibrium. Angles at the particle (between successive forces): $150^{\circ}$ between $\mathbf{F}_2$ and $\mathbf{F}_3$ , $120^{\circ}$ between $\mathbf{F}_1$ and $\mathbf{F}_3$ , so $90^{\circ}$ between $\mathbf{F}_1$ and $\mathbf{F}_2$ (since the angles sum to $360^{\circ}$ ).

Triangle internal angles (opposite each force): $\theta_1 = 30^{\circ}$ , $\theta_2 = 60^{\circ}$ , $\theta_3 = 90^{\circ}$ . Given $|\mathbf{F}_3| = 12$ :

$|\mathbf{F}_1| = 12 \sin 30^{\circ}/\sin 90^{\circ} = 6$ .
$|\mathbf{F}_2| = 12 \sin 60^{\circ}/\sin 90^{\circ} = 6\sqrt{3}$ .

Cosine rule can also help if you need an angle from three given magnitudes.

Three forces in equilibrium $\Leftrightarrow$ closed triangle.
Internal triangle angle = $180^{\circ} -$ external angle at particle.
Sine rule: $P_i/\sin\theta_i$ constant.
Cosine rule if you need an angle from magnitudes.

Lami's theorem

Equivalent to the sine rule on the triangle of forces.

Statement. Three coplanar forces $P_1$ , $P_2$ , $P_3$ acting at a single point in equilibrium satisfy:

$\dfrac{P_1}{\sin\alpha_1} = \dfrac{P_2}{\sin\alpha_2} = \dfrac{P_3}{\sin\alpha_3},$

where $\alpha_i$ is the angle between the OTHER two forces (i.e. the angle 'opposite' force $P_i$ measured at the point of concurrency).

Key insight. The three angles $\alpha_1$ , $\alpha_2$ , $\alpha_3$ are the angles between PAIRS of forces at the particle, NOT the angles in the head-to-tail triangle. They satisfy $\alpha_1 + \alpha_2 + \alpha_3 = 360^{\circ}$ .

The numerical result agrees with the sine rule on the triangle of forces: $\sin(180^{\circ} - \theta) = \sin\theta$ , so the ratios are the same.

When to use Lami's (vs resolving):

Lami's is fast when the question gives three forces with angles between them and you want magnitudes.
Resolving is more general and works for any number of forces.
For Edexcel mark schemes, BOTH are credited. Pick whichever you find clearer.

Worked example (Lami's). Particle in equilibrium under $\mathbf{P} = 20$ N horizontal, $\mathbf{Q}$ at $60^{\circ}$ above horizontal on the opposite side, weight $\mathbf{W}$ vertical. Angles at particle: $\mathbf{P}$ to $\mathbf{Q}$ = $120^{\circ}$ ; $\mathbf{Q}$ to $\mathbf{W}$ = $150^{\circ}$ ; $\mathbf{W}$ to $\mathbf{P}$ = $90^{\circ}$ .

$20/\sin 150^{\circ} = Q/\sin 90^{\circ} = W/\sin 120^{\circ}$ .
$20/0.5 = 40$ . So $Q = 40 \cdot 1 = 40$ N; $W = 40 \cdot \sin 120^{\circ} = 20\sqrt{3} \approx 34.64$ N.

$P_i / \sin\alpha_i$ constant; $\alpha_i$ = angle between OTHER two forces.
Three angles at particle sum to $360^{\circ}$ .
Equivalent to sine rule on triangle of forces.
Resolving works too — pick whichever is clearer.

Equilibrium on a rough inclined plane

Friction adjusts; find the range of $P$ (or $\mu$ ) maintaining equilibrium.

Setup. Particle on a rough incline with an applied force $P$ . Friction $F$ can act EITHER up or down the slope, depending on which way the particle tends to move.

Standard questions:

Find $P$ when about to slide UP. Friction is at its limit, acting DOWN the slope ( $F = \mu R$ ). Resolve along slope.
Find $P$ when about to slide DOWN. Friction is at its limit, acting UP the slope ( $F = \mu R$ ). Resolve along slope. Different sign on the friction term.
Find the range of $\mu$ for which the particle stays at rest given a fixed $P$ . Combine the two cases.

Workflow (for about-to-slide UP):

Force diagram with $R$ perpendicular, $F$ down slope (friction opposing upward motion), $P$ in its specified direction, weight vertical.
Resolve perpendicular: $R = mg\cos\theta + P\sin\alpha$ (or similar, depending on $P$ 's direction).
Friction $= \mu R$ (limiting).
Resolve along slope (up positive): $P\cos\alpha = mg\sin\theta + \mu R$ .
Solve for $P$ .

For about-to-slide DOWN, flip the friction's sign: $P\cos\alpha + \mu R = mg\sin\theta$ (friction now helps support the particle).

Worked example. Box $10$ kg on $25^{\circ}$ rough slope ( $\mu = 0.3$ ) with horizontal applied force $P$ . Find $P_{\max}$ (about to slide up) and $P_{\min}$ (about to slide down).

$R = 10g\cos 25^{\circ} + P\sin 25^{\circ}$ (with $P$ horizontal, pressing into slope).
Case up: $P\cos 25^{\circ} = 10g\sin 25^{\circ} + 0.3 R \Rightarrow P_{\max} \approx 87.3$ N.
Case down: $P\cos 25^{\circ} + 0.3 R = 10g\sin 25^{\circ} \Rightarrow P_{\min} \approx 14.3$ N.
For $14.3 \le P \le 87.3$ , the box stays in equilibrium.

Two cases: about to slide up; about to slide down.
Friction direction depends on tendency of motion.
Range of $P$ for equilibrium: $[P_{\min}, P_{\max}]$ .
Resolve perpendicular FIRST to find $R$ (may depend on $P$ ).

Tension in strings — direction matters

Tension acts along the string, away from the particle.

Tension is the internal pulling force in a stretched string or rope. Key facts:

Acts ALONG the string.
AWAY from the particle (toward the support / other end).
Same value at every point along a LIGHT string (negligible mass).
Cannot be negative — a string can only PULL, not push (unlike a rod).

Setup with two strings supporting a hanging mass. Common scenarios:

Mass hangs from a single string attached to two ceiling points. Two tensions; one is fixed and you might be given the angle.
Mass attached at the end of one string with a horizontal pulling force.

Worked example. Particle of mass $5$ kg held by a horizontal force $F$ and a string making $40^{\circ}$ with the vertical.

Horizontal: $F = T\sin 40^{\circ}$ .
Vertical: $T\cos 40^{\circ} = 5g \Rightarrow T = 5g/\cos 40^{\circ} \approx 63.96$ N.
$F = 63.96 \cdot \sin 40^{\circ} \approx 41.12$ N.

Equivalent shortcut: $F = mg \tan\theta$ (where $\theta$ is the angle from vertical). But always derive from first principles in an exam — earns all the M-marks.

Tension along string, away from particle.
Same tension throughout a light string.
Tension is always non-negative (strings pull, not push).
Standard 'string at angle' setup: $T\cos\theta = W$ , $F = T\sin\theta$ .

How it’s examined

Statics features in every WME01/01 paper as a discrete question ( $6$ - $10$ marks). Common scenarios: mass on two strings, particle on rough incline with applied force, three forces with given angles. Lami's vs resolving: both methods are credited equally. The 'range of $\mu$ ' or 'range of $P$ ' questions are popular ( $10$ - $12$ marks) and require explicit two-case analysis (about to slide up; about to slide down). Examiners deduct for not labelling friction direction in each case and for missing the second case entirely.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WME01/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

Master this Topic

Worked examples, formulae, definitions and the mistakes examiners flag — everything you need to push from a pass to an A*.

Take this whole topic with you

Download a branded revision sheet — worked examples, formulae, definitions and common mistakes for Statics of a particle, ready to print or save as PDF.

Step-by-step worked examples — Statics of a particle

Step-by-step solutions to past-paper-style questions on statics of a particle, written exactly the way a tutor would explain them at the board.

1Particle in equilibrium under two strings (8 marks, M1)
Core• Adapted from WME01/01 January 2024 Q8• statics, equilibrium, two strings
Question
A particle of mass $4\,\text{kg}$ is suspended in equilibrium by two light inextensible strings $OA$ and $OB$ attached to fixed points $A$ and $B$ on a horizontal ceiling. String $OA$ makes an angle of $30^{\circ}$ with the vertical; string $OB$ makes an angle of $60^{\circ}$ with the vertical. Find the tensions in the two strings.
Step-by-step solution
1. Step 1
  Draw a force diagram at the point $O$ (the particle): tension $T_A$ along $OA$ (up and toward $A$ ), tension $T_B$ along $OB$ (up and toward $B$ ), weight $W = 4g = 39.2\,\text{N}$ vertically downward.
2. Step 2
  Place axes horizontal ( $x$ ) and vertical ( $y$ ). $T_A$ makes $30^{\circ}$ with vertical, so components: horizontal $T_A\sin 30^{\circ}$ , vertical $T_A\cos 30^{\circ}$ . Similarly $T_B$ : horizontal $T_B\sin 60^{\circ}$ (in the opposite direction), vertical $T_B\cos 60^{\circ}$ .
3. Step 3
  Resolve horizontally ( $\Sigma F_x = 0$ ):
  $T_A \sin 30^{\circ} = T_B \sin 60^{\circ} \Rightarrow T_A \cdot 0.5 = T_B \cdot \tfrac{\sqrt{3}}{2} \Rightarrow T_A = \sqrt{3}\,T_B \quad \text{(1)}$
4. Step 4
  Resolve vertically ( $\Sigma F_y = 0$ ):
  $T_A \cos 30^{\circ} + T_B \cos 60^{\circ} = 4 g \Rightarrow \tfrac{\sqrt{3}}{2} T_A + \tfrac{1}{2} T_B = 39.2 \quad \text{(2)}$
5. Step 5
  Substitute (1) into (2):
  $\tfrac{\sqrt{3}}{2} \cdot \sqrt{3} T_B + \tfrac{1}{2} T_B = 39.2 \Rightarrow \tfrac{3}{2} T_B + \tfrac{1}{2} T_B = 2 T_B = 39.2$
6. Step 6
  Solve.
  $T_B = 19.6\,\text{N}, \quad T_A = \sqrt{3} \cdot 19.6 \approx 33.95\,\text{N}$
Answer
$T_A \approx 33.95\,\text{N}$ and $T_B = 19.6\,\text{N}$ .
Examiner tip
B1 for diagram. M1 for resolving horizontally. A1 for horizontal equation. M1 for resolving vertically. A1 for vertical equation. M1 for solving simultaneously. A1A1 for both tensions. Alternative: triangle of forces with sine rule, since angles between forces are $30^{\circ}+60^{\circ}=90^{\circ}$ , $90^{\circ}+60^{\circ}=150^{\circ}$ , $90^{\circ}+30^{\circ}=120^{\circ}$ at the vertices.
2Triangle of forces with sine rule (7 marks, M1)
Extended• Adapted from WME01/01 June 2024 Q5• statics, triangle of forces, sine rule
Question
A particle is in equilibrium under three coplanar forces $\mathbf{F}_1$ , $\mathbf{F}_2$ and $\mathbf{F}_3$ . The angle between $\mathbf{F}_2$ and $\mathbf{F}_3$ is $150^{\circ}$ ; the angle between $\mathbf{F}_1$ and $\mathbf{F}_3$ is $120^{\circ}$ . Given $|\mathbf{F}_3| = 12\,\text{N}$ , find $|\mathbf{F}_1|$ and $|\mathbf{F}_2|$ .
Step-by-step solution
1. Step 1
  Three forces in equilibrium can be drawn head-to-tail forming a closed triangle. Use the angles to label the triangle.
2. Step 2
  Angle between $\mathbf{F}_2$ and $\mathbf{F}_3$ is $150^{\circ}$ — that's the external angle. In the triangle (head-to-tail), the internal angle opposite $\mathbf{F}_1$ is $180^{\circ} - 150^{\circ} = 30^{\circ}$ .
3. Step 3
  Similarly, angle between $\mathbf{F}_1$ and $\mathbf{F}_3$ external $= 120^{\circ}$ , so internal angle opposite $\mathbf{F}_2$ is $60^{\circ}$ . The third angle (opposite $\mathbf{F}_3$ ) is $180^{\circ} - 30^{\circ} - 60^{\circ} = 90^{\circ}$ .
4. Step 4
  Apply the sine rule:
  $\dfrac{|\mathbf{F}_1|}{\sin 30^{\circ}} = \dfrac{|\mathbf{F}_2|}{\sin 60^{\circ}} = \dfrac{|\mathbf{F}_3|}{\sin 90^{\circ}} = \dfrac{12}{1} = 12$
5. Step 5
  Compute:
  $|\mathbf{F}_1| = 12 \sin 30^{\circ} = 6\,\text{N}, \quad |\mathbf{F}_2| = 12 \sin 60^{\circ} = 6\sqrt{3} \approx 10.39\,\text{N}$
6. Step 6
  Equivalent statement: Lami's theorem — $|\mathbf{F}_i|/\sin(\text{angle opposite } \mathbf{F}_i) =$ constant, where the angles are the EXTERNAL angles at the vertex (not the head-to-tail triangle angles). Here Lami's gives $|\mathbf{F}_1|/\sin 150^{\circ} = |\mathbf{F}_3|/\sin(\text{angle between } \mathbf{F}_1, \mathbf{F}_2)$ . Same numerical result.
Answer
$|\mathbf{F}_1| = 6\,\text{N}$ and $|\mathbf{F}_2| = 6\sqrt{3} \approx 10.39\,\text{N}$ .
Examiner tip
M1 for sketch of triangle. A1 for internal angles. M1 for sine rule. A1 for setup. A1 for $|\mathbf{F}_1|$ . A1 for $|\mathbf{F}_2|$ . Lami's theorem is equivalent and equally credited. The trickiest step is converting 'angle between forces' to 'angle in the triangle' — always sketch.
3Particle in equilibrium on a rough incline — find $\mu$ range (10 marks, M1)
Extended• Adapted from WME01/01 June 2023 Q6• statics, rough incline, range of friction
Question
A box of mass $10\,\text{kg}$ rests on a rough plane inclined at $25^{\circ}$ to the horizontal. A horizontal force $P$ is applied to the box. The coefficient of friction is $0.3$ . (a) Find the value of $P$ when the box is on the point of sliding up the slope. (b) Find the value of $P$ when the box is on the point of sliding down the slope.
Step-by-step solution
1. Step 1
  Draw force diagram with axes along and perpendicular to the slope. Forces: weight $W = 10g$ vertically down, normal reaction $R$ perpendicular to slope, applied force $P$ horizontally (let's say pushing INTO the slope), friction $F$ along slope (direction depends on which case).
2. Step 2
  Resolve $P$ along/perpendicular to slope: along slope (up positive) $P\cos 25^{\circ}$ ; perpendicular (into slope positive) $P\sin 25^{\circ}$ . Resolve $W$ : along (down positive) $10g\sin 25^{\circ}$ ; perpendicular (into slope) $10g\cos 25^{\circ}$ .
3. Step 3
  Perpendicular equilibrium gives normal reaction:
  $R = 10g\cos 25^{\circ} + P\sin 25^{\circ}$
4. Step 4
  (a) About to slide UP $\Rightarrow$ friction $F = \mu R$ acting DOWN the slope. Along-slope equilibrium:
  $P\cos 25^{\circ} = 10g\sin 25^{\circ} + \mu R$
5. Step 5
  Substitute $R$ and $\mu = 0.3$ .
  $P\cos 25^{\circ} = 10g\sin 25^{\circ} + 0.3(10g\cos 25^{\circ} + P\sin 25^{\circ})$
6. Step 6
  Numerical values: $\cos 25^{\circ} \approx 0.9063$ , $\sin 25^{\circ} \approx 0.4226$ , $10g = 98$ .
  $0.9063 P = 98 \cdot 0.4226 + 0.3(98 \cdot 0.9063 + 0.4226 P)$
7. Step 7
  Simplify the RHS: $41.42 + 0.3(88.82 + 0.4226 P) = 41.42 + 26.65 + 0.1268 P = 68.07 + 0.1268 P$ . So $0.9063 P - 0.1268 P = 68.07 \Rightarrow 0.7795 P = 68.07 \Rightarrow P \approx 87.3\,\text{N}$ .
8. Step 8
  (b) About to slide DOWN $\Rightarrow$ friction acts UP the slope. Along-slope equilibrium:
  $P\cos 25^{\circ} + \mu R = 10g\sin 25^{\circ}$
9. Step 9
  Substitute, $0.9063 P + 0.3(88.82 + 0.4226 P) = 41.42 \Rightarrow 0.9063 P + 26.65 + 0.1268 P = 41.42 \Rightarrow 1.033 P = 14.77 \Rightarrow P \approx 14.3\,\text{N}$ .
Answer
(a) On the point of sliding up: $P \approx 87.3\,\text{N}$ . (b) On the point of sliding down: $P \approx 14.3\,\text{N}$ . For values of $P$ between these limits, the box remains in equilibrium.
Examiner tip
B1 for diagram. M1 for resolving perpendicular. A1 for $R$ . M1 for resolving along slope, case (a). A1 for equation. M1 for solving. A1 for $P \approx 87.3$ . Same again for (b). Cases differ ONLY in the sign of $\mu R$ — many candidates lose case (b) marks by not flipping the friction.
4Light rod / cable — direction of force matters (8 marks, M1)
Extended• Adapted from WME01/01 January 2023 Q7• statics, rod, cable, direction
Question
A particle of mass $5\,\text{kg}$ is held in equilibrium by a horizontal force $F$ and a light string. The string makes an angle of $40^{\circ}$ with the vertical. Find $F$ and the tension in the string.
Step-by-step solution
1. Step 1
  Draw the diagram. Three forces: $W = 5g = 49\,\text{N}$ down, tension $T$ along the string (up and at $40^{\circ}$ from vertical), $F$ horizontal (so it balances the horizontal component of $T$ ).
2. Step 2
  Resolve horizontally:
  $F = T\sin 40^{\circ}$
3. Step 3
  Resolve vertically:
  $T\cos 40^{\circ} = 5g = 49 \Rightarrow T = \dfrac{49}{\cos 40^{\circ}} \approx 63.96\,\text{N}$
4. Step 4
  Substitute back:
  $F = 63.96 \cdot \sin 40^{\circ} \approx 41.12\,\text{N}$
5. Step 5
  Equivalent: $F = 5g \tan 40^{\circ} \approx 49 \cdot 0.8391 \approx 41.12\,\text{N}$ .
Answer
$T \approx 63.96\,\text{N}$ , $F \approx 41.12\,\text{N}$ .
Examiner tip
B1 for diagram. M1 for resolving vertically. A1 for $T$ . M1 for resolving horizontally. A1 for $F$ . The shortcut $F = mg\tan\theta$ for this geometry is widely memorised but should be re-derived from first principles to keep all the M-marks.
5Three coplanar forces using Lami's theorem (8 marks, M1)
Challenge• Adapted from WME01/01 June 2024 Q8• statics, Lami's theorem, three forces
Question
A particle is held in equilibrium by three forces. Force $\mathbf{P}$ has magnitude $20\,\text{N}$ acting horizontally. Force $\mathbf{Q}$ acts at $60^{\circ}$ above the horizontal (on the opposite side from $\mathbf{P}$ ). The third force $\mathbf{W}$ is the weight, vertical. Find the magnitudes of $\mathbf{Q}$ and $\mathbf{W}$ .
Step-by-step solution
1. Step 1
  Sketch the three forces meeting at the particle. Define an explicit geometry: let $\mathbf{P}$ point along $+\mathbf{i}$ , $\mathbf{Q}$ point at $180^{\circ} - 60^{\circ} = 120^{\circ}$ from $+\mathbf{i}$ (i.e. up and to the left), and $\mathbf{W}$ point straight down ( $-\mathbf{j}$ ).
2. Step 2
  Resolve horizontally ( $\Sigma F_x = 0$ ):
  $P + Q\cos 120^{\circ} = 0 \Rightarrow 20 - Q \cdot 0.5 = 0 \Rightarrow Q = 40\,\text{N}$
3. Step 3
  Resolve vertically ( $\Sigma F_y = 0$ ):
  $Q\sin 120^{\circ} - W = 0 \Rightarrow 40 \cdot \tfrac{\sqrt{3}}{2} = W$
4. Step 4
  Compute:
  $W = 20\sqrt{3} \approx 34.64\,\text{N}$
5. Step 5
  Cross-check with Lami's theorem. Angles between adjacent forces: $\mathbf{P}$ and $\mathbf{Q}$ are $120^{\circ}$ apart; $\mathbf{Q}$ and $\mathbf{W}$ are $90^{\circ} + 30^{\circ} = 120^{\circ}$ apart? Actually $\mathbf{Q}$ is at $120^{\circ}$ from $+\mathbf{i}$ , $\mathbf{W}$ at $270^{\circ}$ , so the angle between $\mathbf{Q}$ and $\mathbf{W}$ (going the short way) is $150^{\circ}$ . $\mathbf{W}$ and $\mathbf{P}$ are $90^{\circ}$ apart. Total angles: $120 + 150 + 90 = 360^{\circ}$ ✓.
6. Step 6
  Lami's: $P/\sin(\text{angle between } \mathbf{Q}, \mathbf{W}) = Q/\sin(\text{angle between } \mathbf{W}, \mathbf{P}) = W/\sin(\text{angle between } \mathbf{P}, \mathbf{Q})$ .
  $\dfrac{20}{\sin 150^{\circ}} = \dfrac{Q}{\sin 90^{\circ}} = \dfrac{W}{\sin 120^{\circ}} = \dfrac{20}{0.5} = 40$
7. Step 7
  $Q = 40\sin 90^{\circ} = 40$ , $W = 40\sin 120^{\circ} = 20\sqrt{3}$ . Matches.
Answer
$Q = 40\,\text{N}$ and $W = 20\sqrt{3} \approx 34.64\,\text{N}$ .
Examiner tip
B1 for diagram. M1 for resolving horizontally. A1 for $Q$ . M1 for resolving vertically. A1 for $W$ . Lami's theorem cross-check earns B1. Edexcel mark schemes accept any consistent angle convention.

Key Formulae — Statics of a particle

The formulae you need to memorise for statics of a particle on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Equilibrium of a particle
$\Sigma F_x = 0, \quad \Sigma F_y = 0$
$F_x$
horizontal component of each force (signed)
$F_y$
vertical component of each force (signed)
When to use
For ANY particle in equilibrium under coplanar forces. Resolve each force into the chosen perpendicular axes; set the algebraic sums to zero. NOT in the IAL formula booklet.
Example
Particle suspended by two strings: $T_A\sin 30^{\circ} = T_B\sin 60^{\circ}$ horizontally; $T_A\cos 30^{\circ} + T_B\cos 60^{\circ} = W$ vertically.
Lami's theorem (three forces in equilibrium)
$\dfrac{P_1}{\sin\alpha_1} = \dfrac{P_2}{\sin\alpha_2} = \dfrac{P_3}{\sin\alpha_3}$
$P_i$
magnitude of the $i$ -th force
$\alpha_i$
angle between the OTHER two forces
When to use
Particle in equilibrium under exactly three coplanar concurrent forces. Equivalent to applying the sine rule to the closed triangle of forces. Useful shortcut to avoid resolving. NOT in the formula booklet.
Example
Three forces with angles between them $120^{\circ}$ , $150^{\circ}$ , $90^{\circ}$ : $P/\sin 150^{\circ} = Q/\sin 90^{\circ} = W/\sin 120^{\circ}$ .
Triangle of forces (head-to-tail)
$\text{Three forces in equilibrium } \Leftrightarrow \text{closed triangle when drawn head-to-tail}$
$—$
geometric statement
When to use
Quick visual check that three forces are in equilibrium. Used with the sine rule or cosine rule for magnitudes. Not in the formula booklet but a standard result.
Example
Three forces $P$ , $Q$ , $W$ in equilibrium with internal angles $30^{\circ}$ , $60^{\circ}$ , $90^{\circ}$ : apply sine rule $P/\sin 30^{\circ} = Q/\sin 60^{\circ} = W/\sin 90^{\circ}$ .
Range of $P$ for equilibrium on rough incline
$P_{\text{min}} \le P \le P_{\text{max}}, \text{ with } F = \pm \mu R \text{ at the limits}$
$P$
applied force (e.g. horizontal)
$P_{\text{max}}$
force at which particle is about to slide up
$P_{\text{min}}$
force at which particle is about to slide down
When to use
When a particle on a rough incline is in equilibrium under an applied force, friction can act EITHER way. The range $[P_{\text{min}}, P_{\text{max}}]$ gives all $P$ for which equilibrium is possible.
Example
Box on $25^{\circ}$ slope with $\mu = 0.3$ : $P_{\text{min}} \approx 14.3$ , $P_{\text{max}} \approx 87.3$ N.

Key Definitions and Keywords — Statics of a particle

Definitions to memorise and the exact keywords mark schemes credit for statics of a particle answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Equilibrium
Examiner keyword
A state where the resultant force (and, for a rigid body, the resultant moment) is zero. For a particle: $\Sigma \mathbf{F} = \mathbf{0}$ , equivalent to $\Sigma F_x = 0$ and $\Sigma F_y = 0$ .
Coplanar forces
Examiner keyword
Forces all lying in the same plane. Most M1 problems are coplanar — analysed by resolving into two perpendicular directions in that plane.
Concurrent forces
Forces whose lines of action all pass through a single point. A particle is treated as a point, so all forces on it are concurrent at the particle.
Triangle of forces
Examiner keyword
Three forces are in equilibrium if and only if they form a closed triangle when drawn head-to-tail. The sine rule (or cosine rule) on this triangle gives Lami's theorem.
Limiting equilibrium
Examiner keyword
A state of equilibrium in which friction is at its maximum: $F = \mu R$ . Indicates the particle is on the point of slipping (in a specified direction).

Common Mistakes and Misconceptions — Statics of a particle

The traps other students keep falling into on statics of a particle questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Drawing friction in the SAME direction as the tendency of motion
Why it happens
Confusing 'tendency of motion' with the direction friction acts.
How to avoid it
Friction OPPOSES the tendency of motion. If a force pushes the particle UP a slope, friction acts DOWN the slope. If gravity pulls the particle DOWN, friction acts UP.
✕Using the angle BETWEEN a force and one of the others, instead of the angle OPPOSITE the force in Lami's theorem
WME01/01 examiner reports
Why it happens
Lami's theorem looks deceptively simple.
How to avoid it
In Lami's $P_i / \sin\alpha_i$ , the angle $\alpha_i$ is the angle between the OTHER two forces (the angle 'opposite' force $P_i$ at the point of concurrency).
✕Solving for only one value of $P$ on a rough incline problem
Why it happens
Not realising friction can act either way.
How to avoid it
On a rough incline with applied force, work out BOTH cases: friction up the slope (about to slide down) and friction down the slope (about to slide up). The two values bracket the equilibrium range.
✕Using $R = mg\cos\theta$ when an applied force has a perpendicular component
Why it happens
Memorising the formula without re-deriving.
How to avoid it
On a slope with applied force $P$ at some angle, always RE-DERIVE $R$ by resolving perpendicular to the slope. $R$ may include a $\pm P\sin\theta$ term depending on whether $P$ pushes into or pulls away from the slope.
✕Using $\sin$ instead of $\cos$ (or vice versa) when resolving
Why it happens
Forgetting the geometric convention.
How to avoid it
Mark the angle on the diagram. The component adjacent to the angle is $\cos\theta$ ; the component opposite to the angle is $\sin\theta$ .

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Statics of a particle — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Statics of a particle

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Particle in equilibrium under two strings (8 marks, M1)

2Triangle of forces with sine rule (7 marks, M1)

3Particle in equilibrium on a rough incline — find μ\muμ range (10 marks, M1)

4Light rod / cable — direction of force matters (8 marks, M1)

5Three coplanar forces using Lami's theorem (8 marks, M1)

Equilibrium of a particle

Lami's theorem (three forces in equilibrium)

Triangle of forces (head-to-tail)

Range of PPP for equilibrium on rough incline

Equilibrium

Coplanar forces

Concurrent forces

Triangle of forces

Limiting equilibrium

✕Drawing friction in the SAME direction as the tendency of motion

✕Using the angle BETWEEN a force and one of the others, instead of the angle OPPOSITE the force in Lami's theorem

✕Solving for only one value of PPP on a rough incline problem

✕Using R=mgcos⁡θR = mg\cos\thetaR=mgcosθ when an applied force has a perpendicular component

✕Using sin⁡\sinsin instead of cos⁡\coscos (or vice versa) when resolving

Practice questions

Past paper style quiz

Assess your understanding

Statics of a particle — frequently asked questions

3Particle in equilibrium on a rough incline — find $\mu$ range (10 marks, M1)

Range of $P$ for equilibrium on rough incline

✕Solving for only one value of $P$ on a rough incline problem

✕Using $R = mg\cos\theta$ when an applied force has a perpendicular component

✕Using $\sin$ instead of $\cos$ (or vice versa) when resolving