Equilibrium and Friction are the heart of Statics in Cambridge A-Level Mechanics 1. You will learn how to analyse objects at rest using force balance and moment balance, model friction realistically, and solve problems on slopes, beams, and ladders. Every statics problem comes back to ΣF = 0 and ΣM = 0.
ΣF = 0 (both directions) | F ≤ μN | F = μN (limiting) | Moment = F × d | Lami: F₁/sinα₁ = F₂/sinα₂ = F₃/sinα₃
Learning Objectives
State the conditions for equilibrium: ΣF_x = 0, ΣF_y = 0, ΣM = 0
Apply Lami's theorem to three concurrent coplanar forces
Use the triangle of forces for three-force equilibrium
Model friction using F ≤ μN and identify when F = μN
Distinguish smooth surfaces (μ = 0) from rough surfaces
Solve limiting equilibrium problems for particles and blocks
Set up equilibrium equations for objects on rough inclined planes
Determine the condition tanθ ≤ μ for no slipping on a slope
Calculate moments and apply the principle of moments
Solve beam, see-saw, and ladder problems using moment equilibrium
Equilibrium Conditions
ΣF = 0 in both directions, three-force bodies, Lami's theorem
Friction Model
F ≤ μN, limiting friction, rough vs smooth surfaces
Limiting Equilibrium
On the verge of sliding, F = μN, find μ or unknown forces
Inclined Planes
Slopes with friction, tanθ ≤ μ condition, applied forces
Moments
Turning effect, principle of moments, beams and ladders
Worked Examples
8 fully worked with mark-scheme steps
Common Mistakes
8 errors that regularly cost marks in exams
Exercises + Practice
80+ self-marking questions across all topics
Learn 1 — Equilibrium Conditions
What is Equilibrium?
A particle or rigid body is in equilibrium when it is at rest (or moving at constant velocity with no rotation). For a body in equilibrium, two conditions must hold simultaneously:
Condition 1: ΣF_x = 0 (net horizontal force = 0) Condition 2: ΣF_y = 0 (net vertical force = 0)
Why both conditions?
• Condition 1 ensures the body does not accelerate horizontally.
• Condition 2 ensures the body does not accelerate vertically.
• For a rigid body, a third condition ΣM = 0 is also needed (covered in Learn 5).
• For a particle (point mass), only ΣF = 0 is required — particles cannot rotate.
Setting Up Equilibrium Equations
Method: Step 1: Draw a clear force diagram showing every force acting on the body. Step 2: Choose horizontal (→ positive) and vertical (↑ positive) axes. Step 3: Resolve each force into its horizontal and vertical components. Step 4: Write ΣF_x = 0 and ΣF_y = 0 as two separate equations. Step 5: Solve simultaneously for the unknowns.
Example: Particle Suspended by Two Strings
A particle of weight W hangs from two strings making angles 30° and 45° with the vertical. Find the tensions T₁ and T₂.
Substitute (1) into (2): T₁ (√3/2) + (T₁/√2)(√2/2) = W → T₁(√3/2 + 1/2) = W
T₁ = W / (√3/2 + 1/2) = 2W / (√3 + 1) ≈ 0.732W
Three-Force Equilibrium
When exactly three forces act on a body in equilibrium, two important properties hold:
1. The forces must be concurrent (all three lines of action meet at a single point), OR all three forces must be parallel.
2. Triangle of forces: The three force vectors, drawn head-to-tail, form a closed triangle. This is a direct consequence of ΣF = 0 — the vector sum of three forces being zero means they form a closed triangle.
This allows you to use trigonometry (sine rule, cosine rule) on the force triangle to find unknown magnitudes or directions.
Lami's Theorem
For three concurrent coplanar forces in equilibrium, Lami's theorem states:
F₁ / sin α₁ = F₂ / sin α₂ = F₃ / sin α₃
where α₁ is the angle between F₂ and F₃ (opposite to F₁), α₂ is the angle between F₁ and F₃ (opposite to F₂), and α₃ is the angle between F₁ and F₂ (opposite to F₃).
Key: The angles must be the angles BETWEEN the other two forces (i.e., opposite to each force vector). These three angles sum to 360°.
Example: Three forces 10 N, 8 N, and T N act at a point in equilibrium. The angle between the 10 N and 8 N forces is 120°. By Lami's theorem, T/sin120° = 10/sinα₂.
Lami's theorem is essentially the sine rule applied to the triangle of forces.
Summary of Equilibrium Strategies
Situation
Best Method
Two or more forces at angles
Resolve horizontally and vertically, write ΣF = 0 equations
Exactly three concurrent forces
Lami's theorem or triangle of forces + sine rule
Forces parallel to each other
Direct algebraic sum = 0
Rigid body (not a particle)
ΣF = 0 AND ΣM = 0 about a convenient point
Always check: does the number of unknowns match the number of equations? With ΣF_x = 0 and ΣF_y = 0 you have two equations — you can find at most two unknowns. If there are more unknowns, you need extra equations (moments, or Lami's theorem).
Learn 2 — Friction: The Model
What is Friction?
Friction is a contact force that acts along the surface between two objects, opposing the relative motion (or tendency of motion) between them. In Cambridge A-Level M1, we use a simple but powerful mathematical model:
F ≤ μN (friction is at most μN) F = μN only at limiting equilibrium (on the verge of sliding)
The Coefficient of Friction μ
μ (mu) is the coefficient of friction — a dimensionless number that characterises how rough the surface is.
• μ = 0: perfectly smooth surface (no friction). Labelled "smooth" in exam questions.
• 0 < μ < 1: typical rough surface (e.g., wood on wood ≈ 0.3–0.5).
• μ > 1: very rough surface (possible in theory; rare in M1 problems).
N is the normal reaction force — always perpendicular to the contact surface. Find N from the perpendicular equilibrium equation first.
The Friction Inequality: F ≤ μN
When an object is in static equilibrium on a rough surface:
• Friction F adjusts itself to whatever value is needed to maintain equilibrium.
• BUT friction cannot exceed its maximum value μN.
• If the required friction exceeds μN, the object will slide (equilibrium breaks down).
Key distinction:
• Object at rest but not on verge of slipping: F < μN (friction takes whatever value equilibrium requires)
• Object at rest, on verge of slipping (limiting equilibrium): F = μN
• Object sliding: F = μN (kinetic friction — approximately equal to limiting static friction in M1)
Direction of Friction
Critical rule: Friction always opposes the direction of motion OR the tendency of motion (the direction the object would slide if friction were removed).
• Object tending to slide to the right → friction acts to the LEFT.
• Object on a slope tending to slide down → friction acts UP the slope.
• Object on a slope tending to slide up (due to applied force) → friction acts DOWN the slope.
You must determine the tendency of motion from the context BEFORE assigning a direction to friction. Getting this wrong is a very common source of errors.
Smooth vs Rough Surfaces in Exam Questions
"Smooth surface" → μ = 0, so F = 0 (no friction). Ignore friction entirely. "Rough surface" → μ > 0, friction exists. If the value of μ is not given, it is an unknown to find.
Exam questions will always tell you whether a surface is smooth or rough. If rough, they will either give μ or ask you to find it.
Static vs Kinetic Friction
In real life, static friction (maximum when about to slip) is slightly higher than kinetic friction (during sliding). In Cambridge M1:
• We treat them as equal: F = μN in both limiting equilibrium and during motion.
• This simplification is standard for A-Level purposes.
When setting up friction problems: (1) Find N from the perpendicular equation. (2) Determine the direction of motion/tendency. (3) Write F = μN if limiting, or F = ? if in equilibrium with F unknown. Never assume F = μN unless the question states limiting equilibrium or the object is moving.
Learn 3 — Limiting Equilibrium
Definition
A particle is in limiting equilibrium when it is on the verge of sliding — the friction force has reached its maximum value F = μN. The particle is still stationary (a = 0), but any slight increase in the disturbing force would cause it to move.
At limiting equilibrium: F = μN AND ΣF = 0 (body still stationary)
Setting Up Limiting Equilibrium Problems
Step 1: Draw a force diagram with ALL forces labelled (including friction F and normal reaction N). Step 2: Write the perpendicular equilibrium equation to find N. Step 3: Substitute F = μN (because we're at limiting equilibrium). Step 4: Write the along-surface equilibrium equation (ΣF = 0 in the motion direction). Step 5: Solve for the unknown (μ, an applied force, or a mass).
Finding μ from a Limiting Equilibrium Problem
Example: A 5 kg block rests on a horizontal rough surface. A horizontal force of 24.5 N is just enough to cause slipping. g = 9.8 m s⁻².
Vertical: N = mg = 5 × 9.8 = 49 N
At limiting equilibrium (horizontal): Applied force = F = μN
24.5 = μ × 49
μ = 24.5 / 49 = 0.5
Finding the Force Needed to Cause Limiting Equilibrium
Example: A 4 kg block is on a rough horizontal surface, μ = 0.35, g = 10 m s⁻². What horizontal force P is needed to bring it to the verge of sliding?
Vertical: N = 4 × 10 = 40 N
At limiting equilibrium: F = μN = 0.35 × 40 = 14 N
Horizontal ΣF = 0: P = F = 14 N
Limiting Equilibrium with an Angled Applied Force
Example: A 6 kg block on a rough horizontal surface (μ = 0.4). A force P acts at 20° above horizontal. Find P for limiting equilibrium. g = 10 m s⁻².
Vertical: N + P sin20° = 6 × 10 = 60 → N = 60 − P sin20°
Horizontal at limiting equilibrium: P cos20° = μN = 0.4(60 − P sin20°)
P cos20° = 24 − 0.4P sin20°
P(cos20° + 0.4 sin20°) = 24
P(0.9397 + 0.1368) = 24
P = 24 / 1.0765 = 22.3 N
The "Rough Surface" Clue in Exam Questions
"Rough surface" → friction exists, write F and N as unknowns. "Just about to slide" or "limiting equilibrium" or "on the point of moving" → use F = μN. "Smooth surface" → F = 0, no friction at all.
The phrase "just about to move" is the exam's way of telling you F = μN can be used. Without this phrase, you cannot assume F = μN — friction may be less than its maximum value.
Remember: in limiting equilibrium, the body is STILL stationary (acceleration = 0), so BOTH ΣF = 0 AND F = μN hold simultaneously. Do not apply F = ma — it gives a = 0, which you already know. The two conditions give you your two equations.
Learn 4 — Equilibrium on Inclined Planes
Forces on a Particle on a Rough Slope
For a particle of mass m resting on a rough slope inclined at angle θ to the horizontal, the forces are:
Weight component down slope: mg sinθ Normal reaction (perpendicular to slope): N = mg cosθ Maximum friction (up slope, preventing slip): F = μN = μmg cosθ
Equilibrium along slope: Friction must balance the weight component down the slope:
F = mg sinθ (when in equilibrium)
Equilibrium perpendicular to slope:
N = mg cosθ
For equilibrium to be possible: F ≤ μN
mg sinθ ≤ μ × mg cosθ → sinθ ≤ μ cosθ → tanθ ≤ μ
Condition for No Slipping on a Slope
tanθ ≤ μ (particle does not slip) tanθ = μ (limiting equilibrium — on verge of slipping down)
This is a powerful result: if you know the angle of the slope and the coefficient of friction, you can immediately determine whether a particle will slip.
Example: μ = 0.5 (tan⁻¹(0.5) ≈ 26.6°). Any slope steeper than 26.6° will cause the particle to slide. On exactly 26.6°, it is at limiting equilibrium.
Particle with an Additional Applied Force Along the Slope
Force P applied up the slope (preventing the particle from sliding down):
Along slope (up = positive): P + F − mg sinθ = 0 → P = mg sinθ − F
If at limiting equilibrium (about to slip down): P = mg sinθ − μmg cosθ = mg(sinθ − μcosθ)
Force P applied down the slope (particle about to slip UP due to P being too large):
Along slope (up = positive): F − P − mg sinθ = 0 → P = F − mg sinθ
At limiting equilibrium (about to slip up): P_max = μmg cosθ − mg sinθ = mg(μcosθ − sinθ)
Range of Forces for Equilibrium
Sometimes a question asks for the range of an applied force P for which the particle remains in equilibrium. Friction can act either up or down the slope:
Lower bound (particle about to slip DOWN, friction acts UP):
P_min = mg sinθ − μmg cosθ = mg(sinθ − μcosθ)
(If this is negative, P can be zero and equilibrium still holds.)
If force P acts at angle α to the slope surface:
Perpendicular to slope: N + P sinα = mg cosθ → N = mg cosθ − P sinα
Along slope: P cosα − mg sinθ − F = 0 (particle about to move up)
At limiting equilibrium: P cosα − mg sinθ = μ(mg cosθ − P sinα)
Solve for P: P(cosα + μ sinα) = mg(sinθ + μcosθ)
P = mg(sinθ + μcosθ) / (cosα + μ sinα)
Always resolve perpendicular to the slope FIRST to find N, because N is needed to compute friction F = μN. Also, carefully determine the direction of friction (up or down slope) by identifying whether the particle tends to slip up or down.
Learn 5 — Moments and Turning Effects
What is a Moment?
The moment of a force about a point is its turning effect. It measures how much a force tends to rotate an object about a given pivot.
Moment = Force × perpendicular distance from pivot to line of action of force M = F × d (units: N m)
Sign convention: Anticlockwise moments are positive (+); clockwise moments are negative (−). (This convention is consistent but arbitrary — you can choose your own, but stick with it throughout a problem.)
Key point: The distance d must be the PERPENDICULAR distance from the pivot to the LINE OF ACTION of the force — not just any distance to where the force is applied.
Principle of Moments
For a body in equilibrium, the sum of all moments about ANY point is zero:
ΣM = 0 about any point
Why take moments about a point?
Taking moments about the point where an unknown force acts eliminates that force from the moment equation, making it easier to solve for other unknowns.
Strategy: Choose the pivot at the point of action of the force you want to eliminate from the moments equation.
Beam Problems
Example: A uniform beam AB of mass 10 kg and length 6 m rests horizontally on supports at A and B. A load of 50 N acts at 2 m from A. Find the reactions R_A and R_B. g = 10 m s⁻².
Total forces: R_A + R_B = 50 + 100 = 150 N ...(1) [ΣF_y = 0]
Take moments about A (eliminates R_A):
R_B × 6 = 50 × 2 + 100 × 3 (weight of beam acts at midpoint = 3 m from A)
6R_B = 100 + 300 = 400
R_B = 400/6 = 66.7 N
From (1): R_A = 150 − 66.7 = 83.3 N
Tilting and Overturning
A beam or plank tilts (about one support) when the reaction at the other support becomes zero. At the point of tilting:
• Reaction at one support = 0.
• Take moments about that support to find the maximum load or minimum distance that causes tilting.
Ladder Problems
A ladder of length L and mass m leans against a smooth vertical wall at angle θ to the horizontal. Forces:
• Weight mg downward at midpoint of ladder.
• Normal reaction from wall (horizontal, since wall is smooth) = R_W.
• Normal reaction from rough floor (vertical) = R_F.
• Friction from floor (horizontal) = F (towards wall).
Vertical: R_F = mg
Horizontal: F = R_W
Take moments about base of ladder:
R_W × L sinθ = mg × (L/2) cosθ
R_W = mg cosθ / (2 sinθ) = mg/(2 tanθ)
For equilibrium: F ≤ μR_F → R_W ≤ μmg → mg/(2tanθ) ≤ μmg → tanθ ≥ 1/(2μ)
See-Saw (Lever) Problems
A see-saw of negligible mass pivots at centre. Masses m₁ at distance d₁ and m₂ at distance d₂ on opposite sides:
ΣM = 0: m₁g × d₁ = m₂g × d₂ → m₁d₁ = m₂d₂
This is the law of the lever: heavier mass is closer to pivot, lighter mass is further.
To find reaction forces at supports: use ΣF_y = 0 to get one equation, then take moments about one support to eliminate its reaction and solve for the other. Never take moments about a general point when a support point is available — it makes the algebra much simpler.
Worked Examples
Eight fully worked examples covering all key topics in Equilibrium and Friction. Study each step carefully before moving on.
Example 1 — Three Forces in Equilibrium (Horizontal Particle)
A particle is in equilibrium under three forces: 12 N due East, 5 N due North, and a third force F. Find the magnitude and direction of F.
Example 3 — Find Friction Force and Normal Reaction
A 8 kg block rests on a rough horizontal surface. A horizontal force of 18 N is applied but the block does not move. g = 10 m s⁻². Find the friction force and state whether the surface is rough enough if μ = 0.3.
Step 1 Vertical ΣF = 0: N = mg = 8 × 10 = 80 N
Step 2 Horizontal ΣF = 0 (block stationary): F = 18 N (friction must equal applied force)
Step 3 Maximum friction = μN = 0.3 × 80 = 24 N
Step 4 Since 18 N < 24 N, F ≤ μN is satisfied — the block remains stationary. Equilibrium holds.
Example 4 — Limiting Equilibrium on a Slope: Finding μ
A particle of mass 3 kg rests on a rough slope inclined at 28° to the horizontal and is just about to slide down. Find the coefficient of friction. g = 10 m s⁻².
Step 1 Perpendicular to slope: N = mg cos28° = 30 × 0.8829 = 26.49 N
Step 2 At limiting equilibrium (about to slide down), friction acts up the slope: F = μN
Step 3 Along slope ΣF = 0: F = mg sin28° = 30 × 0.4695 = 14.09 N
Step 4 μ = F/N = 14.09/26.49 = tan28° ≈ 0.532
Example 5 — Equilibrium on Rough Slope with Extra Force
A 5 kg particle is on a rough slope (μ = 0.4) inclined at 35°. A force P acts up the slope. The particle is in limiting equilibrium about to slip UP. Find P. g = 10 m s⁻².
Step 1 Perpendicular: N = 5 × 10 × cos35° = 50 × 0.8192 = 40.96 N
Step 2 Particle about to slip UP → friction acts DOWN the slope: F = μN = 0.4 × 40.96 = 16.38 N
Step 3 Along slope ΣF = 0 (up = positive): P − mg sin35° − F = 0
Step 4 P = 50 × 0.5736 + 16.38 = 28.68 + 16.38 = 45.1 N
Example 6 — Ladder Against a Smooth Wall
A uniform ladder of mass 20 kg and length 5 m leans against a smooth vertical wall, making 65° with the horizontal. The floor is rough with μ = 0.35. Determine if the ladder is in equilibrium. g = 10 m s⁻².
Step 1 Vertical ΣF = 0: R_F = 20 × 10 = 200 N
Step 2 Take moments about base of ladder: R_W × 5sin65° = 200 × 2.5cos65°
Step 4 Required friction = R_W = 46.6 N. Maximum friction = μR_F = 0.35 × 200 = 70 N. Since 46.6 < 70, ladder is in equilibrium.
Example 7 — Beam with Multiple Loads
A uniform beam AB, mass 15 kg, length 8 m, rests on supports at A and at C (2 m from B). Loads: 80 N at 1 m from A, 60 N at 5 m from A. Find reactions at A and C. g = 10 m s⁻².
Step 1 Weight of beam = 150 N acting at midpoint (4 m from A). Support C is at 6 m from A.
Step 4 R_C = 980/6 = 163.3 N. From (1): R_A = 290 − 163.3 = 126.7 N
Example 8 — Equilibrium of a Suspended Particle (Three Strings)
A particle of weight W hangs in equilibrium from a horizontal ceiling by two strings. String 1 makes 40° with the ceiling; String 2 makes 50° with the ceiling. Find tensions T₁ and T₂ in terms of W.
Step 1 Angles with vertical: String 1 at 50° from vertical, String 2 at 40° from vertical.
Step 3 Vertical: T₁ sin50° + T₂ sin40° = W → 0.7660T₁ + 0.6428T₂ = W ...(2)
Step 4 Sub (1): 0.7660T₁ + 0.6428 × 0.839T₁ = W → T₁(0.7660 + 0.5393) = W → T₁ = W/1.305 ≈ 0.766W; T₂ ≈ 0.643W
Common Mistakes
These are the most frequent errors students make in equilibrium and friction questions. Learn to avoid them before your exam.
Mistake 1: Friction Direction — Getting It Wrong
✗ Automatically writing friction up the slope without checking the tendency of motion. If a large force pushes the particle up the slope, friction acts DOWNWARD.
✓ Always ask: "Which way would the particle move if friction were removed?" Friction opposes THAT direction. It is not always up the slope.
Mistake 2: Assuming F = μN When Not at Limiting Equilibrium
✗ Writing F = μN for a stationary particle when the question does not state "just about to move" or "limiting equilibrium". Friction may be less than μN.
✓ Only use F = μN when the question says "limiting equilibrium", "just about to slip/move", or the particle is already sliding. Otherwise F is an unknown satisfying F ≤ μN.
Mistake 3: Wrong Point for Taking Moments
✗ Taking moments about a general point in the middle of a beam, making the equation include all unknowns and very messy to solve.
✓ Take moments about the point where an unknown reaction acts. This eliminates that unknown from the moments equation entirely, leaving only one unknown to solve for.
Mistake 4: Lami's Theorem Angle Assignment
✗ Using the angle between the force and the x-axis in Lami's theorem, e.g. writing F₁/sin30° when 30° is the angle F₁ makes with the horizontal.
✓ In Lami's theorem, each force is divided by the sine of the angle BETWEEN THE OTHER TWO forces (i.e., the angle OPPOSITE to that force). Check all three angles sum to 360°.
Mistake 5: Forgetting the Weight of the Beam Itself
✗ In beam problems, only accounting for the loads applied to the beam and forgetting that a "uniform" beam has its own weight acting at its midpoint.
✓ "Uniform beam of mass m" → weight mg acts at the midpoint (L/2 from each end). Always include this in both the force equation and the moments equation.
Mistake 6: Not Including All Forces in ΣF = 0
✗ Leaving out the normal reaction, or forgetting a string tension, or omitting the weight of the object itself when writing equilibrium equations.
✓ Draw a complete force diagram FIRST, labelling every single force with a letter. Then write ΣF_x = 0 and ΣF_y = 0 using ALL forces from your diagram. No shortcuts.
Mistake 7: Confusing Perpendicular Distance with Slant Distance
✗ Using the distance along a rod or string as the perpendicular distance in a moment calculation when the force is not perpendicular to the rod.
✓ Moment = F × d where d is the perpendicular distance from the pivot to the LINE OF ACTION of F. If the force makes angle α with the rod, use d = L sinα (where L is the length along the rod to the point of application).
Mistake 8: Ignoring the Condition for Equilibrium on a Slope
✗ Setting up an equation with F = μN and solving without checking whether the required friction exceeds μN, or failing to state the range of values that allow equilibrium.
✓ Always verify F ≤ μN. When asked for the range of a force for equilibrium, consider BOTH limiting cases: particle about to slip down (F = μN up slope) and about to slip up (F = μN down slope). These give the minimum and maximum values of the force.
Key Formulas — Equilibrium and Friction
Equilibrium Conditions
ΣF_x = 0 (net horizontal force = 0) ΣF_y = 0 (net vertical force = 0) ΣM = 0 (net moment about any point = 0, for rigid bodies)
Friction Model
F ≤ μN (general: friction does not exceed μN) F = μN (limiting equilibrium or sliding) μ = F/N (to find coefficient of friction)
Lami's Theorem (Three Concurrent Forces)
F₁/sinα₁ = F₂/sinα₂ = F₃/sinα₃ where αᵢ is the angle between the OTHER two forces (opposite to Fᵢ) α₁ + α₂ + α₃ = 360°
Inclined Plane Equilibrium
N = mg cosθ (normal reaction on slope) F = mg sinθ (friction needed to hold particle in equilibrium on slope) Condition for equilibrium: tanθ ≤ μ Limiting equilibrium: μ = tanθ
Applied Force on Slope — Limiting Equilibrium
P_min = mg(sinθ − μcosθ) (particle about to slip DOWN, friction UP) P_max = mg(sinθ + μcosθ) (particle about to slip UP, friction DOWN) (P along slope, directed up)
Moments
Moment = F × d (d = perpendicular distance from pivot to line of action) Anticlockwise positive. ΣM = 0 for equilibrium.
Beam — Reactions at Supports
R_A + R_B = total downward loads + weight of beam Take moments about A: R_B × L = Σ(load × distance from A) Substitute back to find R_A
Ladder Problem
R_W = mg cosθ / (2 sinθ) = mg / (2 tanθ) (reaction from smooth wall) R_F = mg (reaction from floor, vertical) Friction from floor: F = R_W ≤ μR_F Minimum angle for equilibrium: tanθ ≥ 1/(2μ)
Formula
Meaning
When to Use
ΣF_x = 0, ΣF_y = 0
No net force (equilibrium)
Every equilibrium problem
F = μN
Friction at maximum (limiting)
Limiting equilibrium or sliding
F/sinα formula (Lami)
Three concurrent forces
Exactly 3 forces in equilibrium
N = mg cosθ
Normal reaction on slope
Any inclined plane problem
μ = tanθ
Coefficient from slope angle
Limiting equilibrium on slope
M = F × d
Turning moment
Beams, ladders, see-saws
ΣM = 0
No net rotation (equilibrium)
Rigid body problems
Proof Bank
Rigorous derivations of key results in equilibrium and friction. Understanding these proofs deepens your conceptual grasp and helps you adapt to unfamiliar exam questions.
Proof 1 — Lami's Theorem from the Sine Rule on the Triangle of Forces
Given: Three concurrent forces F₁, F₂, F₃ in equilibrium.
Step 1: Since ΣF = 0, the vector sum F₁ + F₂ + F₃ = 0. This means the three force vectors, placed head-to-tail, form a closed triangle.
Step 2: Draw the force triangle. Label the sides as |F₁|, |F₂|, |F₃| opposite to angles A, B, C respectively in the triangle.
Step 3: Apply the sine rule to the force triangle:
|F₁|/sinA = |F₂|/sinB = |F₃|/sinC
Step 4: Relate triangle angles to original angles between forces. If the angle between F₂ and F₃ (i.e., the angle at the vertex of the triangle between these two vectors) is α₁, then the interior angle of the triangle opposite to |F₁| is π − α₁ (supplementary). So sinA = sin(π − α₁) = sinα₁.
This is Lami's Theorem. It is valid only for three concurrent, coplanar forces in equilibrium.
Proof 2 — Condition tanθ ≤ μ for No Slipping on a Slope
Setup: Particle of mass m on rough slope inclined at angle θ to horizontal. Coefficient of friction μ. No additional applied forces.
Step 1: Resolve perpendicular to slope: N − mg cosθ = 0 → N = mg cosθ.
Step 2: The particle tends to slide DOWN the slope (gravity component down slope = mg sinθ). For equilibrium, friction F must act UP the slope.
Step 3: Resolve along slope (up = positive): F − mg sinθ = 0 → F = mg sinθ.
Step 4: For equilibrium, the friction force F must satisfy F ≤ μN:
mg sinθ ≤ μ × mg cosθ
Step 5: Divide both sides by mg cosθ (positive for 0 < θ < 90°):
sinθ/cosθ ≤ μ → tanθ ≤ μ ■
Equality case: tanθ = μ is limiting equilibrium — the particle is on the verge of slipping. If θ increases beyond this, the particle slides.
Proof 3 — Principle of Moments (Derivation of Conditions for Equilibrium)
Setup: A rigid body acted on by forces F₁, F₂, …, Fₙ. For the body to be in equilibrium, it must not translate AND not rotate.
Step 1 (No translation): Newton's second law for the centre of mass: ΣF = ma_cm. For a = 0 (equilibrium), we need ΣF = 0. This gives the two force equations.
Step 2 (No rotation): The angular momentum equation about any fixed point P: ΣM_P = I_P × α. For α = 0 (no angular acceleration), we need ΣM_P = 0.
Step 3 (Independence of pivot): If ΣM_A = 0 about point A and ΣF = 0, then ΣM_B = 0 about any other point B. This follows because:
ΣM_B = ΣM_A + (AB) × ΣF = ΣM_A + 0 = 0 (since ΣF = 0)
Conclusion: For a rigid body in equilibrium, ΣF = 0 AND ΣM = 0 about ANY point. You can choose the most convenient pivot. ■
Visualiser — Interactive Moment Calculator (Beam)
Set up to 4 loads on a beam and choose a pivot position to see the sum of moments. All distances measured from the LEFT end of the beam (total length 10 m).
m
m
m
m
m
Set loads and click "Calculate Moments" to see the sum of moments about your chosen pivot.
Friction Check — Is the Object in Equilibrium?
Check whether a block on a slope remains in equilibrium given the slope angle, coefficient of friction, and any applied force.
Find unknown forces using ΣF = 0. Give numerical answers to 2 d.p. where needed. g = 10 m s⁻².
Exercise 2 — Friction Force Calculations (10 Questions)
Find friction forces, normal reactions, and coefficients of friction. g = 10 m s⁻².
Exercise 3 — Limiting Equilibrium (10 Questions)
Problems involving F = μN at limiting equilibrium. g = 10 m s⁻².
Exercise 4 — Forces on Inclined Planes with Friction (10 Questions)
Equilibrium and limiting equilibrium on rough slopes. g = 10 m s⁻².
Exercise 5 — Moments and Beam Equilibrium (10 Questions)
Calculate moments and find reactions at supports. g = 10 m s⁻².
Practice — 30 Mixed Questions
Mixed questions covering all topics. Get 100% for confetti! g = 10 m s⁻² throughout.
Challenge — 15 Harder Questions
More demanding problems requiring careful reasoning and multi-step solutions. g = 10 m s⁻².
Exam Style Questions (8 Questions)
Cambridge M1-style questions with mark schemes. Attempt fully before revealing the scheme.
Q1 [4 marks]
A particle of mass 3 kg rests in equilibrium under three forces: a force of 20 N due East, a force of 15 N due North, and a third force F. Find the magnitude of F and the bearing on which it acts. (g = 10 m s⁻²)
ΣF_x = 0: F_x = −20 N [M1]
ΣF_y = 0: F_y = −15 N [M1]
|F| = √(400 + 225) = √625 = 25 N [A1]
Bearing = 180° + tan⁻¹(20/15) = 180° + 53.1° = 233.1° (South of West) [A1]
Q2 [5 marks]
A 6 kg block rests on a rough horizontal surface (μ = 0.45). A horizontal force P is applied. Find (i) the friction force when P = 20 N (block stationary), (ii) the value of P at which limiting equilibrium occurs. (g = 10 m s⁻²)
N = mg = 6 × 10 = 60 N [B1]
(i) Block stationary: F = P = 20 N [B1] (friction adjusts; max friction = 0.45 × 60 = 27 N, so equilibrium holds) [M1]
(ii) At limiting equilibrium: F = μN = 0.45 × 60 = 27 N → P = 27 N [M1 A1]
Q3 [5 marks]
A particle of mass 4 kg is on a rough slope inclined at 32° to the horizontal. It is just about to slide down the slope. Find the coefficient of friction and the normal reaction. (g = 10 m s⁻²)
Perpendicular to slope: N = 4 × 10 × cos32° = 40 × 0.8480 = 33.92 N [M1 A1]
Along slope (limiting equilibrium, friction up slope): F = mg sin32° = 40 × 0.5299 = 21.20 N [M1]
μ = F/N = 21.20/33.92 = tan32° ≈ 0.625 [A1 A1]
Q4 [6 marks]
A uniform beam AB has mass 12 kg and length 5 m. It rests horizontally on supports at A and at D, where AD = 3.5 m. A load of 90 N is placed at B. Find the reactions at A and D, and state whether the beam is on the point of tilting. (g = 10 m s⁻²)
Weight of beam = 120 N at 2.5 m from A. [B1]
ΣF_y = 0: R_A + R_D = 90 + 120 = 210 N ...(1) [M1]
Take moments about A: R_D × 3.5 = 120 × 2.5 + 90 × 5 = 300 + 450 = 750 [M1]
R_D = 750/3.5 = 214.3 N [A1]
From (1): R_A = 210 − 214.3 = −4.3 N [A1]
Negative R_A indicates the beam would need to be held down at A — the beam is on the point of tilting about D. [A1]
Q5 [5 marks]
Three forces act on a particle in equilibrium. Force A = 12 N, Force B = 16 N, and Force C. The angle between A and B (measured around the particle) is 130°. Using Lami's theorem, find Force C. (g = 10 m s⁻²)
Angles between forces sum to 360°. Angle between A and B = 130° → angle opposite to C is 130°. [M1]
By Lami's theorem: C/sin130° = 12/sinα_A = 16/sinα_B [M1]
From 12/sinα_A = 16/sinα_B → sinα_B/sinα_A = 16/12 = 4/3. Also α_A + α_B = 360° − 130° = 230°.
Using sinα_B = sin(230° − α_A): Solving numerically, α_A ≈ 107.5°, α_B ≈ 122.5° [M1 A1]
C = 12 × sin130°/sin107.5° = 12 × 0.766/0.9537 ≈ 9.63 N [A1]
Q6 [6 marks]
A particle of mass 5 kg rests on a rough slope (μ = 0.3) inclined at 20° to the horizontal. A force P acts along the slope. Find the range of values of P for which the particle remains in equilibrium. (g = 10 m s⁻²)
N = 5 × 10 × cos20° = 50 × 0.9397 = 46.98 N [M1]
Max friction = μN = 0.3 × 46.98 = 14.09 N [A1]
mg sin20° = 50 × 0.3420 = 17.10 N [B1]
Lower bound (particle about to slip DOWN, F acts up, P also acts up):
P_min + 14.09 = 17.10 → P_min = 3.01 N [M1 A1]
Upper bound (particle about to slip UP, F acts down):
P_max = 17.10 + 14.09 = 31.19 N [A1]
Range: 3.01 N ≤ P ≤ 31.19 N
Q7 [6 marks]
A uniform ladder of mass 18 kg and length 6 m rests with its foot on rough horizontal ground and its top against a smooth vertical wall, making 70° with the ground. A person of mass 72 kg stands 4 m up the ladder. Find the friction and normal reactions at the ground. (g = 10 m s⁻²)
Weight of ladder = 180 N at midpoint (3 m up). Weight of person = 720 N at 4 m up. [B1]
Vertical ΣF = 0: R_ground = 180 + 720 = 900 N [B1]
Take moments about foot of ladder (R_wall × perpendicular distance): [M1]
R_wall × 6sin70° = 180 × 3cos70° + 720 × 4cos70°
R_wall × 5.638 = 180 × 1.026 + 720 × 1.368 = 184.7 + 985.0 = 1169.7 [M1 A1]
R_wall = 1169.7/5.638 = 207.5 N [A1]
Friction at ground = R_wall = 207.5 N (horizontal equilibrium) [A1]
Q8 [7 marks]
A particle of mass 3 kg rests on a rough horizontal surface (μ = 0.5). A force P acts at 25° above the horizontal. Find (i) the value of P for limiting equilibrium, (ii) the minimum force needed to move the particle (hint: optimise angle). (g = 10 m s⁻²)
(i) P at 25°:
Vertical: N = 30 − P sin25° [M1]
Horizontal (limiting): P cos25° = μN = 0.5(30 − P sin25°) [M1]
P cos25° + 0.5P sin25° = 15
P(0.9063 + 0.5 × 0.4226) = 15 → P × 1.117 = 15 [M1 A1]
P = 13.43 N [A1]
(ii) Minimum force (optimum angle α):
P_min is minimised when tan α = μ → α = tan⁻¹(0.5) ≈ 26.6°
P_min = mg sinα_opt / (cosα_opt + μ sinα_opt) ... using P(cosα + μ sinα) = μmg: P_min = μmg/√(1+μ²) = 0.5×30/√1.25 = 15/1.118 ≈ 13.42 N [M1 A1]
Past Paper Questions
5 questions from Cambridge A-Level 9709 Mechanics 1 past papers on Equilibrium and Friction. Attempt fully before revealing the mark scheme.
Past Paper Q1 — 9709/41/M/J/20 Q2 [6 marks]
A particle of mass 5 kg is in equilibrium on a rough plane inclined at 30° to the horizontal, held by a force of magnitude P N acting up and along the plane. The coefficient of friction between the particle and the plane is 0.2. Find the range of values of P. (g = 10 m s⁻²)
N = 5 × 10 × cos30° = 50 × (√3/2) = 43.30 N [M1 A1]
Max friction = 0.2 × 43.30 = 8.66 N [A1]
mg sin30° = 50 × 0.5 = 25 N [B1]
Lower bound (about to slip down, F up slope): P + 8.66 = 25 → P_min = 16.34 N [M1]
Upper bound (about to slip up, F down slope): P = 25 + 8.66 → P_max = 33.66 N [M1] 16.34 N ≤ P ≤ 33.66 N [A1]
Past Paper Q2 — 9709/42/O/N/19 Q3 [6 marks]
A uniform rod AB of mass 8 kg and length 4 m rests horizontally on smooth supports at A and at C, where AC = 3 m. A block of mass m kg is placed on the rod at B. Given that the rod is on the point of tilting about C, find m. (g = 10 m s⁻²)
Rod about to tilt about C → reaction at A = 0. [B1]
Take moments about C: [M1]
Weight of rod acts at midpoint (2 m from A = 1 m to LEFT of C). [B1]
mg × 1 (clockwise about C) = 8g × 1 (also: weight of rod 1m left of C, anticlockwise) [M1]
Clockwise: mg_block × (4−3) = mg × 1 m
Anticlockwise: 8g × (3−2) = 8g × 1 m
mg × 1 = 8g × 1 → m = 8 kg [A1 A1]
Past Paper Q3 — 9709/43/M/J/18 Q4 [7 marks]
A particle of mass 2 kg rests on a rough horizontal surface. A force of magnitude F N acts on the particle at an angle of 40° above the horizontal. The coefficient of friction is 0.35. Find the value of F for limiting equilibrium and the normal reaction. (g = 10 m s⁻²)
Vertical: N + F sin40° = 2 × 10 = 20 → N = 20 − F sin40° [M1]
Horizontal (limiting): F cos40° = μN = 0.35(20 − F sin40°) [M1 A1]
F cos40° = 7 − 0.35F sin40°
F(cos40° + 0.35 sin40°) = 7
F(0.7660 + 0.35 × 0.6428) = 7
F × 0.9910 = 7 [M1]
F = 7.07 N [A1]
N = 20 − 7.07 × sin40° = 20 − 7.07 × 0.6428 = 20 − 4.54 = 15.46 N [M1 A1]
Past Paper Q4 — 9709/41/O/N/17 Q3 [6 marks]
A particle is in equilibrium under three forces: 8 N at bearing 000°, 6 N at bearing 090°, and F N at bearing θ. Find F and θ. (g = 10 m s⁻²)
Force 8 N north: components (0, 8). Force 6 N east: components (6, 0). [B1]
ΣF_x = 0: 6 + F sinθ = 0 → F sinθ = −6 [M1]
ΣF_y = 0: 8 + F cosθ = 0 → F cosθ = −8 [M1]
F² = 6² + 8² = 36 + 64 = 100 → F = 10 N [A1]
tanθ = 6/8 = 0.75 → θ = 36.87° south of west → bearing = 180° + 36.87° = 216.9° [M1 A1]
Past Paper Q5 — 9709/42/M/J/16 Q4 [8 marks]
A uniform plank AB has mass 20 kg and length 8 m. It rests with end A on rough horizontal ground and end B against a smooth vertical wall, making 60° with the ground. A person of mass 60 kg climbs the plank. Find (i) the maximum distance the person can climb before the plank slips, given μ = 0.4 between plank and ground. (g = 10 m s⁻²)
Let person be at distance d from A. Forces: weight 200 N at 4 m (midpoint), person 600 N at d, wall reaction R_W (horizontal at B), floor reaction R_F (vertical), friction F (horizontal at A). [B1]
Vertical: R_F = 200 + 600 = 800 N [B1]
Horizontal: F = R_W [B1]
At limiting equilibrium (about to slip): F = μR_F = 0.4 × 800 = 320 N → R_W = 320 N [M1 A1]
Take moments about A: R_W × 8sin60° = 200 × 4cos60° + 600 × d × cos60° [M1]
320 × 6.928 = 200 × 2 + 600d × 0.5
2216.9 = 400 + 300d [M1]
300d = 1816.9 → d = 6.06 m [A1]