Toppling and Sliding

This lesson covers the conditions under which a rigid body on an inclined surface will slide or topple. Understanding which occurs first is a classic AQA Further Mechanics topic.

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1 Overview

When a block (or other rigid body) sits on a plane inclined at angle $\theta$θ, two failure modes can occur as $\theta$θ increases:

1. Sliding: The component of weight along the slope exceeds the maximum friction force.
2. Toppling: The line of action of the weight falls outside the base of support, creating an overturning moment.

The block will do whichever occurs first (at the smaller angle).

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2 Condition for Sliding

The block slides when:

$$mg\sin\theta > \mu mg\cos\theta$$mgsinθ>μmgcosθ $$\boxed{\tan\theta > \mu}$$tanθ>μ​

The critical angle for sliding: $\theta_s = \arctan(\mu)$θs​=arctan(μ).

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3 Condition for Toppling

Consider a rectangular block of width $2a$2a and height $2h$2h, standing on the inclined plane. The centre of mass is at height $h$h above the base and $a$a from the lower edge.

The block topples about its lower edge when the vertical line through the centre of mass passes beyond this edge. The critical condition is:

$$\boxed{\tan\theta > \frac{a}{h}}$$tanθ>ha​​

Derivation:

The centre of mass is at a perpendicular distance $a$a from the lower face and $h$h from the base. Taking moments about the lower edge of the base:

• The weight acts vertically through the centre of mass.
• The perpendicular distance from the lower edge to the line of action of the weight is $a\cos\theta - h\sin\theta$acosθ−hsinθ.

The block topples when this distance becomes zero:

$a\cos\theta = h\sin\theta$acosθ=hsinθ

$\tan\theta = a/h$tanθ=a/h

The critical angle for toppling: $\theta_t = \arctan(a/h)$θt​=arctan(a/h).

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4 Which Occurs First?

Condition	What happens first
$\mu < a/h$μ<a/h	Sliding occurs first ($\theta_s < \theta_t$θs​<θt​)
$\mu > a/h$μ>a/h	Toppling occurs first ($\theta_t < \theta_s$θt​<θs​)
$\mu = a/h$μ=a/h	Sliding and toppling occur simultaneously

Key insight: Tall, thin objects (small $a/h$a/h) topple more easily. Short, wide objects (large $a/h$a/h) slide more easily.

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Example 1

A uniform rectangular block is 0.6 m wide and 1.0 m tall. It stands on a rough slope with $\mu = 0.4$μ=0.4. Does it slide or topple first?

Solution

$a = 0.3$a=0.3 m, $h = 0.5$h=0.5 m.

Topple when $\tan\theta = a/h = 0.3/0.5 = 0.6$tanθ=a/h=0.3/0.5=0.6, so $\theta_t = \arctan(0.6) = 31.0^\circ$θt​=arctan(0.6)=31.0∘.

Slide when $\tan\theta = \mu = 0.4$tanθ=μ=0.4, so $\theta_s = \arctan(0.4) = 21.8^\circ$θs​=arctan(0.4)=21.8∘.

Since $\theta_s < \theta_t$θs​<θt​, the block slides first.

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Example 2

The same block but with $\mu = 0.8$μ=0.8. Does it slide or topple?

Solution

$\theta_s = \arctan(0.8) = 38.7^\circ$θs​=arctan(0.8)=38.7∘.