Equilibrium of Rigid Bodies | 9231 Further Mechanics

1. Centre of Mass — Concept and Symmetry

The centre of mass (CoM) of a rigid body is the single point through which gravity effectively acts. For a rigid body in equilibrium, the weight acts as a single downward force at the CoM.

💡 Key Principle

For a uniform body, the CoM lies at the geometric centre (on all axes of symmetry). You must be able to identify this by inspection before calculating anything.

Standard Results from MF19 Formula Booklet

These are given in the MF19 booklet — you must know how to apply them, not prove them.

Shape	Description	Distance of CoM	From
△ Triangular lamina	Uniform, any triangle	`1/3 × median length`	Each vertex (intersection of medians)
▽ Solid cone/pyramid	Uniform, height h	`h/4`	Base
◗ Solid hemisphere	Uniform, radius r	`3r/8`	Flat face (centre)
◑ Hemispherical shell	Uniform, radius r	`r/2`	Flat face (centre)
◻ Uniform rod/rectangle	Uniform	Midpoint	Either end / edge
⌒ Circular arc (radius r, angle 2α)	Uniform wire	`r sinα / α`	Centre of circle
◔ Sector (radius r, angle 2α)	Uniform lamina	`2r sinα / (3α)`	Centre of circle

2. Composite Bodies — Centre of Mass

A composite body is built from simpler shapes joined together. We treat each part as a particle at its own CoM and use the weighted average formula.

⭐ The Composite CoM Formula

(Σm) x̄= Σ(mᵢ xᵢ)x-coordinate

(Σm) ȳ= Σ(mᵢ yᵢ)y-coordinate

For uniform laminas: mass ∝ area. For uniform solids: mass ∝ volume. A hole is treated as a negative mass.

Method: L-Shaped Lamina

Split the L-shape into two rectangles. Find CoM of each, then apply the formula.

L-shaped lamina split into rectangles A (blue) and B (gold). Green cross = composite CoM.

Let the lamina have uniform density. Taking corner as origin:

Areas (mass proportional)

m_A= 6a × 2a = 12a²

m_B= 3a × 4a = 12a²

Total m= 24a²

CoM of each part

G_A= (3a, 5a) from bottom-left

G_B= (1.5a, 2a) from bottom-left

Composite CoM

x̄= (12a²×3a + 12a²×1.5a) / 24a² = 2.25a

ȳ= (12a²×5a + 12a²×2a) / 24a² = 3.5a

3. Equilibrium Conditions

A rigid body under coplanar forces is in equilibrium if and only if both conditions hold simultaneously:

⭐ The Two Equilibrium Conditions

Condition 1Vector sum of all forces = 0ΣFx = 0, ΣFy = 0

Condition 2Sum of moments about ANY point = 0ΣM = 0

The converse also holds: if a body is in equilibrium, both conditions are satisfied. Choose your moment point wisely — taking moments about an unknown force eliminates it from the equation.

The Moment of a Force

The moment of a force about a point is the product of the force and the perpendicular distance from the point to the line of action of the force.

Moment= Force × perpendicular distanceunits: N m

SignAnticlockwise = positive (convention)

📐 Exam Technique — Taking Moments

Take moments about the point where the most unknowns act — this eliminates them in one step
For a ladder problem: take moments about the base to find the wall reaction first
For a beam: take moments about one end to find the reaction at the other end
Always state your pivot point: "Taking moments about A..."

4. Friction in Equilibrium Problems

When a body is on the verge of sliding, friction reaches its maximum value. The relationship between friction F, normal reaction R, and coefficient of friction μ is:

F ≤ μRin general (not sliding)

F = μRon the point of sliding (limiting equilibrium)

⚠ Critical Distinction

A body can fail to remain in equilibrium in two ways — by sliding or by toppling. These require different conditions and different analysis.

Slides when F > μR — friction is insufficient to prevent motion
Topples when the line of action of the weight passes outside the base — moments are unbalanced

Tilting and Toppling

A body on a surface topples before it slides if the overturning moment exceeds the restoring moment. At the point of toppling, the body pivots about its lowest edge and the reaction at the opposite base point becomes zero.

✓ Toppling Condition

The body topples when the vertical line through the CoM passes beyond the pivot point (the edge it would rotate about). Equivalently, the moment of the weight about the pivot exceeds the moment of the restoring forces.

The Ladder Problem — Classic 9231 Setup

Uniform ladder of weight W and length L, leaning at angle α. Forces: R (normal floor), F (friction floor), S (normal wall), W (weight at midpoint).

Standard Ladder Analysis (smooth wall, rough floor)

Resolve →F = Shorizontal equilibrium

Resolve ↑R = Wvertical equilibrium

Moments (base)S × L sinα = W × (L/2) cosαmoments about base point

∴ S= W cosα / (2 sinα) = W/(2tanα)

Limiting F = μR:μ = F/R = S/W = 1/(2tanα)minimum μ

Example 1 Centre of Mass of a Composite Solid ★★☆ Challenging

A uniform solid cylinder of radius r and height 3r is joined at its base to a solid hemisphere of the same radius r and the same uniform density. Find the distance of the centre of mass of the composite solid from the flat base of the hemisphere.

1

Masses (proportional to volume)

V_cyl= πr² × 3r = 3πr³

V_hemi= (2/3)πr³

Total V= 3πr³ + (2/3)πr³ = (11/3)πr³

2

Position of each CoM (from flat base of hemisphere)

Taking origin at the flat base of the hemisphere (the join):

CoM hemi= −(3r/8)below the join (into hemisphere)

CoM cyl= +(3r/2)above the join (midpoint of cylinder)

3

Apply composite formula — taking origin at flat base of hemisphere

(11/3)πr³ × ȳ= 3πr³ × (3r/2) + (2/3)πr³ × (−3r/8)

= 9πr⁴/2 − 6πr⁴/24 = 9πr⁴/2 − πr⁴/4

= 18πr⁴/4 − πr⁴/4 = 17πr⁴/4

ȳ= (17πr⁴/4) / ((11/3)πr³) = (17r/4) × (3/11)

ȳ= 51r/44 above flat base of hemisphere

Example 2 Ladder in Equilibrium — Finding Minimum Friction ★★☆ Challenging

A uniform ladder of mass 20 kg and length 6 m rests with its upper end against a smooth vertical wall and its lower end on rough horizontal ground. The ladder makes an angle of 65° with the horizontal. A person of mass 60 kg stands on the ladder at a point 4 m from the base. Find (a) the normal reactions at each end, (b) the minimum coefficient of friction at the ground. (g = 9.8 m/s²)

1

Draw diagram and label forces

Forces: R (normal ground ↑), F (friction ground →), S (normal wall ←), W_ladder = 196 N at 3 m, W_person = 588 N at 4 m.

2

Part (a) — Take moments about base (eliminates R and F)

Perpendicular distances: for a force at distance d along ladder at angle 65°, horizontal perpendicular distance = d cos65°.

S × 6 sin65°= 196 × 3 cos65° + 588 × 4 cos65°

S × 5.438= 196 × 1.268 + 588 × 1.690

S × 5.438= 248.5 + 994.2 = 1242.7

S= 1242.7 / 5.438 = 228.5 N

3

Resolve forces

Resolve ↑: R= 196 + 588 = 784 N

Resolve →: F= S = 228.5 N

4

Part (b) — Minimum coefficient of friction

μ_min= F/R = 228.5/784 = 0.291

This is the minimum μ needed to prevent the ladder from sliding at the floor.

Example 3 Toppling vs Sliding — Which Happens First? ★★★ A* Level

A uniform rectangular block of width 2a and height 6a rests on a rough horizontal surface (μ = 0.4). A horizontal force P is applied at the top of the block. Determine whether the block slides or topples first, and find the critical value of P in terms of the weight W.

1

Condition for sliding

The block slides when the applied horizontal force P exceeds maximum friction:

Slides when P≥ μR = μW = 0.4W

P_slide= 0.4W

2

Condition for toppling

The block topples about its leading bottom edge when the moment of P about that edge exceeds the restoring moment of W:

Toppling momentP × 6a (height of force)

Restoring momentW × a (half-width to edge)

Topples whenP × 6a ≥ W × a

P_topple= W/6 ≈ 0.167W

3

Conclusion

Since P_topple (0.167W) < P_slide (0.4W), the block topples first at P = W/6.

The tall, narrow shape makes toppling far easier than sliding. A wide, low block would slide first.

Example 4 Suspended Lamina — Finding the Angle of Tilt ★★☆ Challenging

A uniform L-shaped lamina is formed by removing a square of side a from the corner of a square of side 3a. The lamina is freely suspended from corner A (top-left of original square). Find the angle that the side AB makes with the vertical when the lamina hangs in equilibrium.

1

Find CoM of lamina

Full square area = 9a², removed square area = a², remaining = 8a².

x̄= [9a² × (3a/2) − a² × (5a/2)] / 8a²

= [27a³/2 − 5a³/2] / 8a² = 22a³/(2 × 8a²) = 11a/8

ȳ= [9a² × (3a/2) − a² × (5a/2)] / 8a² = 11a/8

By symmetry of the L-shape (equal amounts removed from both sides), x̄ = ȳ in this case.

2

When suspended from A, CoM hangs directly below A

Corner A is at (0, 3a) taking bottom-left as origin. CoM is at (11a/8, 11a/8).

The vertical through A passes through the CoM. The horizontal distance from A to CoM is 11a/8, vertical distance is 3a − 11a/8 = 13a/8.

tan θ= horizontal/vertical = (11a/8)/(13a/8) = 11/13

θ= arctan(11/13) = 40.2°

The side AB (originally vertical left edge) makes 40.2° with the vertical.

Interactive Moment & Equilibrium Tool

Place up to three forces on a beam and see whether it is in equilibrium. Watch the moment balance update in real time.

⚖ Beam Moment Calculator

Beam length = 10 m. Pivot is at position 0. Adjust force magnitudes and positions. Positive = upward, Negative = downward.

Force 1 (N) 50 N

Position 1 (m) 2.0 m

Force 2 (N) -80 N

Position 2 (m) 5.0 m

Force 3 (N) 60 N

Position 3 (m) 8.0 m

—Net Force (N)

—Net Moment (N·m)

—Required Reaction at Pivot

Practice Questions

Question 1 — Centre of Mass of Composite Lamina

A uniform lamina consists of a rectangle of width 6 cm and height 4 cm with a square of side 2 cm removed from one corner. Taking axes along the two longer sides of the rectangle, find the coordinates of the centre of mass of the lamina.

Treat the removed square as a negative mass. Full rectangle area = 24 cm², removed square = 4 cm². Apply the composite formula with the removed piece having negative mass.

✓ Solution

Full rectangle: area = 24, CoM at (3, 2). Removed square: area = 4, CoM at (5, 3) — placed at top-right corner.

Remaining m= 24 − 4 = 20

x̄= (24×3 − 4×5)/20 = (72−20)/20 = 52/20 = 2.6 cm

ȳ= (24×2 − 4×3)/20 = (48−12)/20 = 36/20 = 1.8 cm

Question 2 — Uniform Beam on Two Supports

A uniform beam AB of mass 12 kg and length 5 m rests horizontally on two supports, one at A and one at a point C, 1 m from B. A mass of 8 kg is placed at B. Find the reactions at A and C. (g = 9.8 m/s²)

Take moments about A to find R_C, then resolve vertically to find R_A. Weight of beam acts at its midpoint (2.5 m from A). Support C is at 4 m from A.

✓ Solution

W_beam = 117.6 N at 2.5 m, W_load = 78.4 N at 5 m. C is at 4 m from A.

Moments (A)R_C × 4 = 117.6 × 2.5 + 78.4 × 5

= 294 + 392 = 686

R_C= 686/4 = 171.5 N

R_A= 117.6 + 78.4 − 171.5 = 24.5 N

Question 3 — Ladder Problem

A uniform ladder of mass 15 kg and length 8 m leans against a smooth vertical wall, with its foot on rough ground. The angle of inclination is 60°. Find the minimum coefficient of friction necessary to prevent slipping. If a man of mass 75 kg wishes to climb to the top of the ladder, what is the minimum coefficient of friction required? (g = 9.8 m/s²)

For part 1: take moments about base, forces are W_ladder at 4m, S at top. For part 2: add man's weight at 8m from base. In both cases μ_min = F/R = S/R.

✓ Solution

Ladder alone (W = 147 N at 4m):

S × 8sin60°= 147 × 4cos60° = 147 × 2 = 294

S= 294/(8 × 0.866) = 42.4 N

R = W= 147 N

μ_min= 42.4/147 = 0.289

With man (extra 735 N at 8m):

S × 6.928= 294 + 735 × 4cos60° = 294 + 1470 = 1764

S= 1764/6.928 = 254.7 N

R= 147 + 735 = 882 N

μ_min= 254.7/882 = 0.289

Interesting result: same μ! This is because for a uniform ladder, μ = 1/(2tanα) regardless of a person's weight at the top.

Question 4 — Suspended Body / Equilibrium Angle

A solid uniform hemisphere of radius r and a solid uniform cylinder of the same radius r and height h are joined together at their flat circular faces. The composite solid is placed with the curved surface of the hemisphere resting on a rough horizontal surface. Find the condition on h for the composite solid to rest in equilibrium without toppling.

Find the CoM of the composite solid from the flat base of the hemisphere (the junction). The solid is stable if the CoM is below the centre of the hemisphere (height 0 in hemisphere coordinates), meaning CoM must be below the centre of curvature for stable equilibrium on a curved surface. Actually, for resting on curved surface: stable if CoM is below the centre of the sphere.

✓ Solution

Volumes: V_hemi = (2/3)πr³, V_cyl = πr²h. CoM positions from flat join (upward positive):

CoM_hemi= −3r/8 (below join, into hemisphere)

CoM_cyl= +h/2 (above join)

ȳ= [πr²h(h/2) + (2/3)πr³(−3r/8)] / [πr²h + (2/3)πr³]

= [h²/2 − r²/4] / [h + 2r/3] (dividing by πr²)

The solid is stable when resting on the hemisphere if the CoM is below the centre of the hemisphere sphere (i.e., ȳ < 0 — CoM is on the hemisphere side of the join). This requires:

h²/2 − r²/4< 0

h²< r²/2

h< r/√2 = r√2/2

When h < r/√2, the CoM lies within the hemisphere half — the solid rests stably. When h ≥ r/√2 it topples.

Formula Reference Sheet

Complete reference for Equilibrium of Rigid Bodies — Cambridge 9231 P3, Section 3.2.

Standard CoM Positions (MF19)

Triangular lamina1/3 from each side along median

Solid cone/pyramidh/4 from base

Solid hemisphere3r/8 from flat face

Hemispherical shellr/2 from flat face

Circular arc (angle 2α)r sinα / α from centre

Sector (angle 2α)2r sinα / (3α) from centre

Composite Body Formula

(Σm) x̄ = Σ(mᵢ xᵢ)

(Σm) ȳ = Σ(mᵢ yᵢ)

Hole = negative masssubtract area/volume

Uniform lamina: m ∝ area

Uniform solid: m ∝ volume

Equilibrium Conditions

ΣFx = 0horizontal forces

ΣFy = 0vertical forces

ΣM = 0 (any point)moments

Moment = F × d⊥perpendicular distance

Friction

F ≤ μRgeneral

F = μRlimiting equilibrium

tan λ = μangle of friction λ

Tilting / Toppling

Topples when CoM passes beyond pivot edge

At topple point: R at far edge = 0

Slides when F > μR

Compare P_slide vs P_topple

Suspended Body

CoM hangs directly below pivot

tan θ = horiz / vert distanceof CoM from pivot

Use geometry carefully

📋 Cambridge Exam Strategy — Rigid Bodies

Always draw a complete force diagram before writing any equations — label every force including direction
Choose your moment point to eliminate unknowns — take moments about a point where 2+ unknown forces act
For composite body CoM, set up a clear table with shape, mass, x-position, y-position — examiners follow your working
When a body is about to topple, one reaction becomes zero — state this explicitly
For suspended bodies, remember: the CoM is directly below the suspension point — this is the key geometric constraint
State units (N, m, N·m) throughout — marks are lost for missing units in moment calculations