Equilibrium of a laminaEdexcel International A Level Further Maths: Revision notes

A lamina suspended from a point $P$ is acted on by two forces only: its weight $W$ at the centre of mass $G$ and the reaction at $P$. For equilibrium these must be equal, opposite and have the same line of action. So the reaction is vertical, equal to $W$, and $G$ hangs vertically below $P$. The weight acts at $G$ whatever the shape, so the lamina simply turns until $G$ is directly below the point of suspension. Once you know where $G$ is, the angle that any edge makes with the vertical comes from a right-angled triangle that has $PG$ as a vertical side.

Example: a uniform rectangle $ABCD$ with $AB=0.6$ and $BC=0.4$ hangs from $A$. $G$ is at $0.3$ along $AB$ and $0.2$ along $AD$. $AG$ is vertical, so $AB$ makes angle $\theta$ with the vertical where $\tan\theta=\frac{0.2}{0.3}$, $\theta=33.7^\circ$. If it hangs from the midpoint of $AB$, $G$ is directly below that point when $AB$ is horizontal. For composite laminas, first find $G$ by moments (the previous topic), then use its coordinates: if $G$ is $\bar{x}$ along one edge and $\bar{y}$ from it, the angle between that edge and the vertical $AG$ has $\tan\theta=\frac{\bar{y}}{\bar{x}}$.

A lamina may be free to rotate about a smooth fixed horizontal axis (a pivot). If a particle is attached, the lamina plus particle is a composite body, so find the combined centre of mass: $\left(M+m\right)\bar{x}=M x_G+m x_P$. It then hangs with the combined centre of mass directly below the pivot. If instead a horizontal force $F$ holds a lamina of weight $W$ at rest, take moments about the pivot (the reaction there has no moment): $F\times(\text{vertical distance from pivot})=W\times(\text{horizontal distance of }G\text{ from pivot})$.

A lamina standing with one edge on a rough plane inclined at $\alpha$ can fail in two ways. It slides if the friction needed exceeds the limit: resolving gives $F=W\sin\alpha$ and $R=W\cos\alpha$, so sliding happens if $\tan\alpha>\mu$. It topples about the lowest corner when $G$ is vertically above that corner, because the weight then has no moment to keep it down the plane. For $G$ at distance $d$ along the edge from the lower corner and height $h$ above the plane, it is about to topple when $\tan\alpha=\frac dh$. Whether it slides or topples first depends on which of $\frac dh$ and $\mu$ is smaller.

Method for any lamina equilibrium question: (1) find the centre of mass $G$ of the lamina, including any attached particles; (2) decide which condition applies (suspension: $G$ below the point; pivot with a force: moments; incline: $G$ above the corner for toppling, $F\le\mu R$ for sliding); (3) draw a right-angled triangle with $G$ and the vertical; (4) calculate and give angles to $0.1^\circ$. Take $g=9.8$ m s$^{-2}$ if the weight is needed.