Equilibrium of Rigid Bodies

Why two conditions? A particle is in equilibrium when the forces balance ($\sum F = 0$). A rigid body can also spin, so it needs a second condition: the forces must produce no net turning effect either. Both must hold.

Condition 1 — no resultant force. The vector sum of all forces is zero. In practice you resolve in two perpendicular directions (usually horizontal and vertical) and set each sum to zero: $\sum F_x = 0, \qquad \sum F_y = 0$

Condition 2 — no resultant moment. The algebraic sum of the moments of all forces about any point is zero: $\sum M = 0 \quad \text{(about any chosen point)}$

Because it works about any point, you may choose the most convenient point — usually one that kills an unknown force (see the last section).

The standard recipe.

1. Draw a clear free-body diagram: mark every force (weight, normal reactions, friction, hinge components).
2. Resolve horizontally ($\sum F_x = 0$).
3. Resolve vertically ($\sum F_y = 0$).
4. Take moments about a well-chosen point ($\sum M = 0$).
5. Solve the resulting equations for the unknowns.

Cambridge tip. Always draw the free-body diagram first and label every force. Most lost marks in this topic trace back to a missing force (often a friction or a hinge component) on the diagram.

Rough contact. A rough surface provides friction $F$ that opposes sliding. Friction is not fixed — it grows to whatever is needed (up to a limit) to keep the body still: $F \leq \mu R$ where $R$ is the normal reaction and $\mu$ the coefficient of friction. On the point of slipping (limiting equilibrium) friction reaches its maximum: $F = \mu R$

The classic ladder. A uniform ladder of weight $W$ and length $2a$ leans at angle $\theta$ to the horizontal against a smooth vertical wall, with its foot on rough horizontal ground. The forces are:

• weight $W$ down, acting at the centre (distance $a$ along the ladder),
• a horizontal normal reaction $S$ from the smooth wall (smooth $\Rightarrow$ no friction at the wall),
• a vertical normal reaction $R$ and a horizontal friction $F$ at the foot (pointing towards the wall, opposing the tendency to slide out).

Resolve and take moments.

• Vertically: $R = W$.
• Horizontally: $F = S$.
• Moments about the foot (this kills $R$ and $F$, since both act at the foot): $S \,(2a\sin\theta) = W\,(a\cos\theta) \;\Rightarrow\; S = \tfrac{1}{2}W\cot\theta$

On the point of slipping, $F = \mu R$, i.e. $S = \mu W$. Substituting $S = \tfrac{1}{2}W\cot\theta$: $\tfrac{1}{2}W\cot\theta = \mu W \;\Rightarrow\; \cot\theta = 2\mu \;\Rightarrow\; \boxed{\tan\theta = \frac{1}{2\mu}}$

This is the standard result: a uniform ladder against a smooth wall on rough ground is about to slip when $\tan\theta = \dfrac{1}{2\mu}$.

Cambridge tip. Point the ground friction towards the wall: the ladder tends to slide its foot out, so friction (which opposes motion) acts inwards. Getting this direction right is essential before you write $F = \mu R$.

As you tilt a body, or push it harder, it will eventually move — but how? Two distinct failures compete:

Sliding (point of slipping). The body keeps full contact but the friction needed exceeds the maximum available, so it slides. This happens when $F = \mu R.$

Toppling (point of overturning). The body rotates about the edge that stays in contact (the pivot edge). As it begins to tip, the normal reaction migrates to that edge. On the point of toppling, the weight's line of action passes exactly through the pivot edge — beyond that, the weight's moment topples it.

Comparing the two. Suppose a horizontal force (or a tilt angle) is slowly increased. Compute the value at which sliding would begin and the value at which toppling would begin. Whichever happens at the smaller value occurs first.

• For a body of weight $W$, base width $2b$, with the weight at height $h$, a horizontal force $P$ at height $h$ topples it when $Ph = Wb$, i.e. $P = \dfrac{Wb}{h}$.
• The same force slides it when $P = \mu R = \mu W$.
• If $\mu W < \dfrac{Wb}{h}$ (i.e. $\mu < \dfrac{b}{h}$) it slides first; if $\mu > \dfrac{b}{h}$ it topples first.

A useful tilt result: a block of base width $2b$ and height $2h$ on a rough plane topples (about its lower edge) when the plane is tilted so that $\tan\alpha = \dfrac{b}{h}$ — the angle at which the vertical through $G$ reaches the lower edge.

Cambridge tip. "On the point of toppling" $\Rightarrow$ put the entire normal reaction at the pivot edge and make the weight line pass through it; "on the point of sliding" $\Rightarrow$ set $F = \mu R$. Decide which test you are applying before writing equations.

The moment condition $\sum M = 0$ holds about any point, and that freedom is your most powerful tool. A force has zero moment about any point on its own line of action, so:

Take moments about the point where the unknown forces you don't want act.

Why it helps. A hinge (or the foot of a ladder) can exert a force with two unknown components $X$ and $Y$. If you take moments about the hinge itself, both $X$ and $Y$ pass through that point and contribute zero moment — the moment equation then contains only the forces you do want, often giving one unknown directly.

Hinge forces in general. A hinge or pivot can push or pull in any direction, so represent its reaction by two perpendicular components $X$ (horizontal) and $Y$ (vertical). You then have three equations — $\sum F_x = 0$, $\sum F_y = 0$, $\sum M = 0$ — for the unknowns. Taking moments about the hinge removes $X$ and $Y$ from that equation.

The general strategy.

1. Identify the unknown you can do without (often a two-component hinge reaction).
2. Take moments about the point where those forces act — they drop out.
3. Solve for the remaining unknown, then resolve to recover the rest.

Cambridge tip. If a question gives a hinge and a supporting force, your first moment equation should almost always be taken about the hinge. It turns a three-unknown system into a one-line answer.