Equilibrium, sliding and topplingAQA A-Level Further Maths: Revision notes

A rigid body is in equilibrium when two conditions hold at once:

• the resultant force is zero, so resolve in two perpendicular directions (e.g. horizontally and vertically);
• the resultant moment is zero about any point. The moment of a force about a point is $F\times d$, where $d$ is the perpendicular distance from the point to the line of action of the force. You may take moments about any point, so choose the point where most unknown forces act: they then have zero moment and drop out. Example: a uniform rod $AB$ of length $4$ m and weight $6g$ is supported at $A$ and at $C$, with $AC=3$ m. Moments about $A$: $3T_C=6g\times2$, so $T_C=4g$. Resolving vertically then gives $T_A=6g-4g=2g$.

A couple is a pair of equal and opposite parallel forces whose lines of action are different. Its resultant force is zero, so it only turns the body. Its moment is $F\times d$, where $d$ is the perpendicular distance between the two forces, and this is the same about every point. In an equilibrium problem a couple is just another moment: include its value in the moment equation, but never in the force equation. A body acted on by a couple can only be kept in equilibrium by an equal and opposite moment, for example from another couple. Example: two forces of $8$ N, $0.5$ m apart, act in opposite directions on a rod. The couple has moment $8\times0.5=4$ N m.

For a ladder against a smooth wall, the wall exerts only a force $R$ perpendicular to the wall. The ground, if rough, exerts a normal reaction $N$ and friction $F$.

1. Draw all forces: weight at the centre of a uniform ladder, $R$, $N$ and $F$.
2. Resolve vertically: $N=W$. Resolve horizontally: $F=R$.
3. Take moments about the foot to find $R$.
4. Use $F\le\mu N$. Example: a $5$ m ladder of mass $20$ kg has its foot $3$ m from the wall, so it reaches $4$ m up. Moments about the foot: $4R=20g\times1.5$, so $R=73.5$ N. Then $F=73.5$ N and $N=196$ N, so $\mu\ge\frac38$.

A body on a rough surface can fail in two ways.

• Sliding: the friction needed exceeds the maximum available, $F>\mu N$. On a slope of angle $\theta$, a block is on the point of sliding when $\tan\theta=\mu$.
• Toppling: the body is on the point of turning about a lower edge. Then the normal reaction acts at that edge, so the weight (or the resultant of the applied forces) has zero moment about it. For a uniform cuboid of width $w$ and height $h$ on a slope, the weight passes through the lower edge when $\tan\theta=\frac{w/2}{h/2}=\frac wh$. To decide which happens first, find both critical conditions and compare: the smaller angle (or force) occurs first. A block topples first on a slope if $\frac wh<\mu$. Example: $w=0.30$, $h=0.80$, $\mu=0.5$: toppling at $\tan\theta=0.375$ ($20.6^\circ$), sliding at $0.5$ ($26.6^\circ$), so it topples first.

A body suspended freely from a point is in equilibrium only when its centre of mass is vertically below the point of suspension. Use this to find the angle at which a lamina or a loaded rod hangs. Example: a rectangle with $AB=0.9$ and $AD=0.6$ hangs from $A$. The centre of mass is $0.45$ along $AB$ and $0.3$ along $AD$, so $AB$ makes an angle $\tan^{-1}\frac{0.3}{0.45}=33.7^\circ$ with the vertical. If a particle is added, find the new centre of mass using a weighted mean ($\bar x=\frac{\sum m_ix_i}{\sum m_i}$), then repeat the trigonometry. When the body is held at a different angle by an external force, use moments about the pivot with the weight acting through the centre of mass.