Equilibrium, toppling and slidingEdexcel A-Level Further Maths: Revision notes

A rigid body under coplanar forces is in equilibrium when the resultant force is zero and the resultant moment about any point is zero. In practice: resolve in two perpendicular directions, then take moments about a well-chosen point (one where several unknown forces act, so they drop out of the equation). $\sum F_x=0,\quad\sum F_y=0,\quad\sum M_O=0.$ The weight $W=mg$ acts at the centre of mass $G$. The moment of a force about a point is force $\times$ perpendicular distance from the point to the line of action. If only three forces act, their lines of action must meet at one point (or be parallel).

A body freely suspended from a point $P$ can only be in equilibrium if the resultant moment about $P$ is zero. The weight acts at $G$, so $G$ must lie vertically below $P$ (the force at $P$ is then vertical and equal to the weight). Example: a uniform rectangle $ABCD$ with $AB=12$, $BC=8$, suspended from $A$. $G$ is the midpoint of $AC$, so $AC$ is vertical and $AB$ makes an angle $\theta$ with the vertical where $\tan\theta=\frac{8}{12}$, so $\theta=33.7^\circ$. If an extra particle of weight $w$ is attached at a point, take moments about the suspension point: the total moment of the weights must be zero, which fixes the angle at which the body hangs.

A block on a plane has three forces: weight $W=mg$ at $G$, the normal reaction $R$ and friction $F$ from the plane. Friction obeys $F\le\mu R$, with $F=\mu R$ in limiting equilibrium. On a plane inclined at $\alpha$: along the slope $F=mg\sin\alpha$, perpendicular to the slope $R=mg\cos\alpha$. Unlike a particle, the normal reaction does not act at a fixed point: its line of action is found by taking moments. $R$ and $F$ together act through a point of the contact surface. If this point would have to lie outside the base, the block cannot be in equilibrium and it topples.

A block is on the point of toppling when it is about to turn about an edge. At that instant the whole contact force ($R$ and $F$) acts at that edge, so taking moments about the edge eliminates both. Block on a slope, width $w$ along the slope and height $h$: the weight acts at the centre, $\frac w2$ along and $\frac h2$ up. Moments about the lower edge give $mg\sin\alpha\times\frac h2=mg\cos\alpha\times\frac w2$, so $\tan\alpha=\frac{w}{h}.$ Block on horizontal ground, force $P$ at height $x$: toppling about the far bottom edge needs $Px=mg\times\frac w2$.

A body slides when the force down the slope (or applied force) reaches the limiting friction. Compare the two conditions and see which is met first. On a slope: sliding at $\tan\alpha=\mu$, toppling at $\tan\alpha=\frac wh$. If $\mu<\frac wh$ the block slides first; if $\mu>\frac wh$ it topples first. Worked example: 30 kg block, base 0.5 m, height 1.2 m, $\mu=0.6$, horizontal $P$ at height 1.0 m. Sliding: $P=0.6\times294=176.4$ N. Toppling: $P(1.0)=294\times0.25$, so $P=73.5$ N. The smaller value happens first, so the block topples.

Draw a clear force diagram with $W$, $R$, $F$ and any applied force, marking the distances to $G$. Use $g=9.8$ unless told otherwise, and give answers to 3 significant figures. For a rod against a smooth wall, the wall force is perpendicular to the wall; against a rough wall add a friction term. Explain conclusions in words, for example: the toppling force is smaller, so the block topples first.