Toppling and Sliding

Put a tall box on a plank and slowly tilt it: at some angle it either slides down the plank or topples over its lower edge — and which happens first depends on a clean competition between the coefficient of friction and the shape of the body. This lesson derives both failure conditions from first principles (force balance for sliding, moments for toppling), shows how to decide which occurs first by comparing $\tan\theta$ with $\mu$ and with $a/h$ , extends to non-uniform bodies and to applied forces, and develops the exam technique of always testing both modes. It is a perennial favourite in the Mechanics option because it rewards careful reasoning about the line of action of the weight.

1 Where this sits in AQA 7367

This is Paper 3, the Mechanics option (7367/3M). It is the natural sequel to centre of mass — toppling is entirely about where the weight's line of action falls relative to the base, which is fixed by $\bar G$ . The content sits at the heart of rigid-body statics: resolving forces, taking moments, and applying the friction inequality $F \le \mu R$ . It is heavily AO2/AO3 (Paper 3 weights these at 25%/35%): you must model the situation, decide the failure mode, and justify your conclusion. Command words include "show that", "determine which occurs first", and "find the least/greatest".

2 Core theory — the two failure conditions

A rigid body rests on a plane inclined at angle $\theta$ . As $\theta$ increases from zero, equilibrium can fail in two distinct ways, and the body does whichever happens at the smaller angle. The two modes are governed by entirely different physics — sliding by a force balance (friction versus the along-slope weight), toppling by a moment balance (the restoring versus overturning effect of the weight about an edge) — so they must be analysed separately and then compared. The whole art of the topic is to compute each threshold cleanly and then make the comparison explicit.

Sliding

Resolve along and perpendicular to the slope. The normal reaction balances the perpendicular weight component, $R = mg\cos\theta$ ; the friction $F$ (up the slope) balances the along-slope weight component $mg\sin\theta$ until it reaches its limiting value $F_{\max} = \mu R$ . The body is on the point of sliding when $mg\sin\theta = \mu mg\cos\theta$ , i.e.

\boxed{\ \tan\theta = \mu\ } \quad\Rightarrow\quad \theta_s = \arctan\mu .

For $\theta > \theta_s$ the required friction exceeds what is available, and the body slides. Notice the mass cancels — the sliding angle depends only on $\mu$ .

Toppling

Model the body as a uniform rectangular block of width $2a$ and height $2h$ ; its centre of mass $G$ sits at the centre, a perpendicular distance $a$ from the lower face and $h$ above the base. The weight $mg$ acts vertically downward through $G$ . The block is in danger of overturning about its lower edge $O$ . Take moments about $O$ : the perpendicular distance from $O$ to the vertical line of action of the weight is

d = a\cos\theta - h\sin\theta .

While $d > 0$ the weight's line of action falls inside the base (on the uphill side of $O$ ) and the weight provides a restoring moment. The block is on the point of toppling when $d = 0$ :

a\cos\theta = h\sin\theta \quad\Rightarrow\quad \boxed{\ \tan\theta = \frac{a}{h}\ } \quad\Rightarrow\quad \theta_t = \arctan\frac{a}{h} .

For $\theta > \theta_t$ the line of action passes beyond $O$ , the moment becomes overturning, and the block topples. Again the mass cancels — toppling depends only on the aspect ratio $a/h$ .

It is worth seeing why $d = a\cos\theta - h\sin\theta$ is the right distance. Set up axes along the slope. The lower edge $O$ and the centre of mass $G$ are separated by a displacement with components $a$ (along the base, towards $G$ ) and $h$ (perpendicular to the base). The weight is vertical; resolving the $O$ -to- $G$ displacement into the horizontal direction (the direction perpendicular to the vertical weight) gives $a\cos\theta$ from the along-base part and $-h\sin\theta$ from the perpendicular part, hence the horizontal offset $a\cos\theta - h\sin\theta$ of $G$ from $O$ . Written as a single sinusoid via the $R$ -formula, $d = \sqrt{a^2 + h^2}\,\cos(\theta + \varphi)$ where $\tan\varphi = h/a$ ; this is zero when $\theta + \varphi = \tfrac{\pi}{2}$ , i.e. $\theta = \tfrac{\pi}{2} - \varphi$ , giving $\tan\theta = \cot\varphi = a/h$ once more — a neat cross-check using A-Level trigonometry.

Which one wins?

Both critical angles are increasing functions of their argument ( $\arctan$ is increasing), so comparing the angles is the same as comparing $\mu$ with $a/h$ :

Condition	Smaller critical angle	What happens first
$\mu < a/h$	$\theta_s < \theta_t$	slides first
$\mu > a/h$	$\theta_t < \theta_s$	topples first
$\mu = a/h$	$\theta_s = \theta_t$	on the verge of both at once

The physical reading: tall, narrow bodies (small $a/h$ ) topple easily; short, wide bodies (large $a/h$ ) and rough surfaces (large $\mu$ ) favour sliding. This single comparison, $\mu$ versus $a/h$ , is the crux of almost every question in the topic.

3 Worked examples with mark scheme

Example 1 — slide or topple? (rough)

A uniform block is $0.6\,\text{m}$ wide and $1.0\,\text{m}$ tall on a rough slope with $\mu = 0.4$ . Which happens first as the slope is tilted?

Here $a = 0.3\,\text{m}$ (half-width) and $h = 0.5\,\text{m}$ (half-height). (M1)

\theta_t = \arctan\frac{a}{h} = \arctan\frac{0.3}{0.5} = \arctan 0.6 \approx 31.0^\circ, \quad (\textbf{M1 A1})

\theta_s = \arctan\mu = \arctan 0.4 \approx 21.8^\circ. \quad (\textbf{A1})

Since $\theta_s < \theta_t$ (equivalently $\mu = 0.4 < a/h = 0.6$ ), the block slides first. (A1) (M1 identify $a,h$ as half-dimensions; M1 toppling angle; A1; A1 sliding angle; A1 conclusion with comparison.)

Example 2 — the same block on a rougher slope

The same block, now with $\mu = 0.8$ . Which happens first?

$\theta_t = 31.0^\circ$ as before; $\theta_s = \arctan 0.8 \approx 38.7^\circ$ . (M1 A1) Now $\theta_t < \theta_s$ (i.e. $a/h = 0.6 < \mu = 0.8$ ), so the block topples first. (A1) (M1 for the new sliding angle; A1; A1 conclusion. The shape is unchanged, so only $\theta_s$ moves — increasing $\mu$ eventually makes toppling the first failure.)

Example 3 — toppling under a horizontal push

A uniform cube of side $0.4\,\text{m}$ and mass $10\,\text{kg}$ sits on a rough horizontal floor. A horizontal force $P$ is applied along the top edge. Find the least $P$ to topple it, and the least $\mu$ for it to topple rather than slide. Take $g = 9.8\,\text{m s}^{-2}$ .

Take moments about the far bottom edge (the pivot the cube would rotate about). The weight $mg$ acts through the centre, a horizontal distance $0.2\,\text{m}$ from the pivot (restoring); $P$ acts at height $0.4\,\text{m}$ (overturning). On the point of toppling: (M1)

P(0.4) = mg(0.2) = 10(9.8)(0.2) = 19.6 \quad\Rightarrow\quad P = \frac{19.6}{0.4} = 49\,\text{N}. \quad (\textbf{M1 A1})

For it to topple before sliding, friction must hold the base while $P$ reaches $49\,\text{N}$ : we need $P \le F_{\max} = \mu mg$ , i.e. (M1)

49 \le \mu (10)(9.8) = 98\mu \quad\Rightarrow\quad \mu \ge 0.5. \quad (\textbf{A1})

(M1 for taking moments about the correct edge; M1 the moment equation; A1 $P = 49\,\text{N}$ ; M1 the friction inequality; A1 $\mu \ge 0.5$ . The pivot is the edge away from the push — getting the pivot right is the whole battle.)

Example 4 — a non-uniform block

A block of width $0.4\,\text{m}$ and height $0.8\,\text{m}$ has its centre of mass $0.15\,\text{m}$ from the (lower) left face and $0.35\,\text{m}$ above the base. It rests on a slope with the lower edge on the left. Find the toppling angle, and state the least $\mu$ that would ensure it topples rather than slides.

Because the body is non-uniform, use the actual centre-of-mass distances: $a = 0.15\,\text{m}$ (horizontal distance from $G$ to the lower edge) and $h = 0.35\,\text{m}$ (height of $G$ above the base). (M1)

\tan\theta_t = \frac{a}{h} = \frac{0.15}{0.35} = \frac{3}{7} \approx 0.4286 \ \Rightarrow\ \theta_t \approx 23.2^\circ. \quad (\textbf{M1 A1})

To topple before sliding we need $\theta_t < \theta_s$ , i.e. $a/h < \mu$ , so $\mu > \tfrac37 \approx 0.43$ . (A1) (M1 for reading off the actual $a,h$ — NOT the half-dimensions, since the block is non-uniform; M1 A1 for the angle; A1 for the $\mu$ condition. The trap is to halve $0.4$ and $0.8$ ; for a non-uniform body the given centre-of-mass position governs everything.)

4 Specimen-style exam question

(specimen-style — not from any past paper)

A uniform rectangular crate of width $0.8\,\text{m}$ , height $1.2\,\text{m}$ and mass $40\,\text{kg}$ stands on rough horizontal ground ( $\mu = 0.35$ ). A horizontal rope is attached at height $h_P$ above the ground and pulled with steadily increasing tension $T$ . (a) Find the tension at which the crate would begin to slide. (b) If the rope is attached at the top $(h_P = 1.2\,\text{m})$ , find the tension at which it would begin to topple. (c) State, with a reason, what actually happens first, and find the greatest $h_P$ for which the crate slides rather than topples. Take $g = 9.8\,\text{m s}^{-2}$ .

Model solution. The weight is $mg = 40(9.8) = 392\,\text{N}$ , and on horizontal ground $R = mg = 392\,\text{N}$ .

(a) Sliding begins when $T = F_{\max} = \mu R = 0.35(392) = 137.2\,\text{N}$ . (Independent of $h_P$ .)

(b) Toppling is about the far bottom edge; the weight acts $0.4\,\text{m}$ (half-width) from it, $T$ acts at $1.2\,\text{m}$ . On the point of toppling:

T(1.2) = mg(0.4) = 392(0.4) = 156.8 \ \Rightarrow\ T = \frac{156.8}{1.2} \approx 130.7\,\text{N}.

(c) With the rope at the top, the toppling tension ( $130.7\,\text{N}$ ) is less than the sliding tension ( $137.2\,\text{N}$ ), so the crate topples first. To make it slide first we need the toppling tension to exceed $137.2\,\text{N}$ ; since $T_{\text{topple}} = \dfrac{mg(0.4)}{h_P} = \dfrac{156.8}{h_P}$ , set

\frac{156.8}{h_P} \ge 137.2 \ \Rightarrow\ h_P \le \frac{156.8}{137.2} \approx 1.143\,\text{m}.

So if the rope is attached at any height up to about $1.14\,\text{m}$ the crate slides first; above that it topples first. The marks hinge on taking moments about the correct edge and on recognising that a lower attachment point favours sliding.

Example 5 — overhanging stacked blocks

Two identical uniform blocks, each of length $L$ , are stacked. The lower block rests on a table with its right end at the table edge. The upper block is placed on the lower one and pushed to the right. How far can the upper block overhang the lower one before it topples, and how far can the pair's combined overhang reach the table edge?

The upper block topples about the right edge of the lower block when its own centre of mass (at its mid-length) passes beyond that edge. So the maximum overhang of the upper block over the lower is

\frac{L}{2}

Toppling and Sliding

Toppling and Sliding

1 Where this sits in AQA 7367

2 Core theory — the two failure conditions

Sliding

Toppling

Which one wins?

3 Worked examples with mark scheme

Example 1 — slide or topple? (rough)

Example 2 — the same block on a rougher slope

Example 3 — toppling under a horizontal push

Example 4 — a non-uniform block

4 Specimen-style exam question

Example 5 — overhanging stacked blocks

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