Equilibrium of rigid bodiesEdexcel International A Level Further Maths: Revision notes

A rigid body is in equilibrium when (1) the resultant force is zero, so components in two perpendicular directions each sum to zero, and (2) the sum of the moments about any point is zero. The moment of a force about a point is the force multiplied by the perpendicular distance from the point to its line of action. Take moments about the point where the most unknown forces act, so those forces drop out. A weight acts at the centre of mass: the midpoint of a uniform rod. If three non-parallel forces keep a body in equilibrium, their lines of action meet at a point, which gives a quick check on the direction of an unknown reaction.

A body freely suspended from a point hangs with its centre of mass vertically below the point, because the weight and the support force must have no moment about the point. Use this to find the angle at which a side hangs from the vertical. Example: a uniform right-angled triangle $ABC$ with $B$ the right angle has centroid a third of the way from each perpendicular side. The angle at which $AB$ hangs follows from $\tan\theta=\frac{\text{perpendicular distance of }G\text{ from }AB}{\text{distance of }G\text{'s foot from }A\text{ along }AB}$. If extra weights are added, find the new centre of mass of the combined body first.

A hinge exerts a force of unknown magnitude and direction, so show it as a horizontal component $X$ and a vertical component $Y$. Take moments about the hinge to remove both. Example: a uniform rod of weight $W=40$ N and length $L$ is hinged at $A$ and held at $45^\circ$ to the vertical by a horizontal force $P$ at $B$. Moments about $A$: $P(L\cos45^\circ)=40\left(\frac L2\sin45^\circ\right)$, so $P=20$ N. Resolving: $X=20$, $Y=40$. The hinge force has magnitude $\sqrt{20^2+40^2}=44.7$ N at $\tan^{-1}0.5=26.6^\circ$ to the vertical. In general the rod at angle $\theta$ to the vertical has $P=\frac W2\tan\theta$.

A body on a plane has a normal reaction $R$ and friction $F\le\mu R$. It slides when $F$ reaches $\mu R$, which on an incline occurs at $\tan\theta=\mu$. It topples when the vertical through the centre of mass passes through the lowest edge of the base, since the reaction can then act no further down the slope. Example: a uniform block with a base of width $0.3$ m along the slope and height $0.4$ m. The centre of mass is $0.2$ m above the base and $0.15$ m from the lower edge, so it topples at $\tan\theta=\frac{0.15}{0.2}=0.75$ ($36.9^\circ$). It topples before sliding if $\mu>0.75$, and slides first if $\mu<0.75$. On a horizontal plane under a horizontal push, the same two outcomes apply.

A smooth surface exerts a reaction perpendicular to itself and no friction; a rough surface exerts both a normal reaction and friction. For a ladder with its foot on rough ground and top against a smooth wall: vertically $R=W$; horizontally $F=S$; moments about the foot give $S$. For a ladder of weight $W$ at angle $\alpha$ to the horizontal, $S\times L\sin\alpha=W\times\frac L2\cos\alpha$, so $S=\frac{W}{2\tan\alpha}$, and the foot does not slip if $\mu\ge\frac{S}{R}=\frac{1}{2\tan\alpha}$. At $\alpha=70^\circ$, $\mu\ge0.182$. A person of weight $w$ at distance $x$ from the foot adds $wx\cos\alpha$ to the right side of the moment equation and $w$ to $R$.