Worked examples — Equilibrium of rigid bodies — translational + rotational

---

The scenario matrix

Every equilibrium problem is one (or a mix) of these cells. We will hit all of them.

Cell	What makes it different	Example
C1 Sign of torque	one force turns clockwise, one counter-clockwise	Ex 1
C2 Force through pivot	its moment arm is zero → it vanishes	Ex 1, Ex 4
C3 Angled force	must split into perpendicular component ($F\sin \theta$)	Ex 2
C4 Zero / degenerate input	force passes through pivot, or angle = 0/90°	Ex 3
C5 Limiting behaviour	let an angle → 0 or → 90°, watch answer blow up or vanish	Ex 5
C6 Couple (zero net force, non-zero torque)	force balance holds, body still spins	Ex 6
C7 Friction unknown (inequality)	solve for a minimum $\mu$	Ex 7
C8 Real-world word problem	translate words → forces → equations	Ex 8
C9 Exam twist (find the pivot reaction, not just the balance)	two support reactions, solve both	Ex 9

Prerequisite refreshers if any word feels shaky: Torque, Centre of Mass, Couple and Moment of a Couple, Static Friction, Centre of Gravity vs Centre of Mass, Newton's Laws of Motion.

---

Example 1 — The unequal seesaw (cells C1, C2)

Forecast: Guess before reading on — will the right force be bigger or smaller than $30\text{ N}$? (It sits farther out…)

Look at the figure: the two red arrows are the applied forces; the black double-arrows are the moment arms $0.4\text{ m}$ and $0.6\text{ m}$ measured along the rod from $O$. Because the rod is horizontal and the forces are vertical, these along-the-rod distances are the perpendicular distances $d$ — no angle correction needed.

Step 1 — Choose the pivot at $O$. Why this step? The support at $O$ pushes up on the rod, but that push acts at $O$, so its perpendicular distance to $O$ is $d=0$ → its torque is zero. Choosing $O$ makes the unknown support force vanish (cell C2).

Step 2 — Write each torque with its sign. Why this step? Torque magnitude is $\tau =Fd$ where $d$ is the perpendicular distance from pivot to the force's line. The left force turns the rod counter-clockwise (positive, +), the right force turns it clockwise (negative, −) — this is cell C1.

$\sum {\tau }_O=0:(30)(0.4)-(F)(0.6)=0$

Step 3 — Solve. $12=0.6F⟹F=20\text{ N}.$

Verify: Left twist $=30\times 0.4=12\text{N⋅m}$; right twist $=20\times 0.6=12\text{N⋅m}$. Equal and opposite → net torque zero. ✓ The far force is smaller, matching intuition: more distance, less force needed.

---

Example 2 — A force pulling at an angle (cell C3)

Forecast: Only part of that $40\text{ N}$ actually twists the rod. Which part — and is it the sine or cosine piece?

Look at the figure: the red arrow is the full pull $T$. The black double-arrow along the rod is $r=0.5\text{ m}$ — the distance from $O$ to the point where the rope grips. The short black upward arrow is the perpendicular slice $T\sin 30^{\circ }$; the dotted horizontal piece $T\cos 30^{\circ }$ points along the rod and does nothing. The little arc marks the $30^{\circ }$ angle between rope and rod.

Step 1 — Split $\vec{T}$ into two pieces. Why this step? A force pointing partly along the rod cannot twist it — pulling straight along a door's edge never opens it. Only the piece perpendicular to the rod turns it. That perpendicular piece is $T\sin \theta$; the along-the-rod piece is $T\cos \theta$ and is wasted for turning (cell C3).

Step 2 — Write the torque. Why this step? Torque = (perpendicular force) × (distance along rod) $=(T\sin \theta )r$. Equivalently, the moment arm is $d=r\sin \theta$, so $\tau =Td=Tr\sin \theta$ — same thing.

$\tau =Tr\sin \theta =(40)(0.5)\sin 30^{\circ }.$

Step 3 — Compute. Since $\sin 30^{\circ }=0.5$: $\tau =40\times 0.5\times 0.5=10\text{N⋅m}.$

Verify: Units $=\text{N}\times \text{m}=\text{N⋅m}$ ✓. Sanity: if the rope pulled straight up ($\theta =90^{\circ }$, $\sin =1$) we'd get the full $40\times 0.5=20\text{N⋅m}$ — twice as much, which is right because all of the force would then be perpendicular.

---

Example 3 — Degenerate case: force through the pivot (cell C4)

Forecast: Will there be any twist at all?

Step 1 — Find the perpendicular distance. Why this step? The rope's line of action passes straight through $O$, so the perpendicular distance from $O$ to that line is $d=0$ (cell C4 — a degenerate input).

Step 2 — Apply the formula. $\tau =Fd=40\times 0=0\text{N⋅m}.$

Verify: Equivalently, the angle between $\vec{r}$ (from $O$ to the pull point) and $\vec{F}$ is $0^{\circ }$, and $\sin 0^{\circ }=0$, so $\tau =Fr\sin 0^{\circ }=0$. ✓ Lesson: any force whose line passes through the pivot contributes zero torque — this is why clever pivot choices work.

---

Example 4 — Balanced plank, weight at the centre of mass (cells C1, C2)

Forecast: The plank's own weight — does it help the person, or fight them? Guess before solving.

Look at the figure: the black triangle is the support, $1.5\text{ m}$ from the left end. The black arrow at the centre is the plank's own weight $W$; the red arrow at the left end is the person $P$. The two double-arrows are the moment arms from the support: the person is $1.5\text{ m}$ to the left, the weight $0.5\text{ m}$ to the right — they sit on opposite sides, which is exactly why they can balance.

Step 1 — Locate the plank's weight. Why this step? For a uniform body, the whole weight acts at the centre of mass — the geometric centre, here $2\text{ m}$ from the left end (see Centre of Gravity vs Centre of Mass). Putting $W$ anywhere else gives a wrong moment arm (Mistake D from the parent).

Step 2 — Choose the pivot at the support. Why this step? The support's upward normal force acts at the support → moment arm $d=0$, so it disappears (cell C2). Only $W$ and $P$ remain.

Step 3 — Perpendicular distances from the support. Why this step? The support is $1.5\text{ m}$ from the left end, and forces are vertical while the plank is horizontal, so along-the-plank distances are the moment arms $d$.

• Plank weight $W$: centre at $2\text{ m}$ from the left; support at $1.5\text{ m}$ → $d=0.5\text{ m}$ to the right of the support → turns clockwise.
• Person $P$: at the left end ($0\text{ m}$ from left) → $d=1.5\text{ m}$ to the left of the support → turns counter-clockwise.

Step 4 — Balance the torques. Why this step? Counter-clockwise (person) must equal clockwise (plank weight) — cell C1. $P(1.5)=W(0.5)⟹P(1.5)=60\times 0.5=30.$

Step 5 — Solve. $P=\frac{30}{1.5}=20\text{N}.$

Verify: Person torque $=20\times 1.5=30\text{N⋅m}$ CCW; plank torque $=60\times 0.5=30\text{N⋅m}$ CW → they cancel. ✓ The plank's own weight fights the person (opposite sense), so the person's weight is exactly what's needed to balance it.

---

Example 5 — Limiting behaviour of the ladder (cell C5)

Forecast: A near-vertical ladder — does it need lots of friction or almost none?

Look at the figure: the red curve is ${\mu }_{\min }=\frac{1}{2}\cot \theta$ plotted against the ladder angle $\theta$. Notice it dives to $0$ on the right (steep ladder) and shoots up towards infinity on the left (flat ladder). The two black dots mark the answers we compute for $30^{\circ }$ and $60^{\circ }$.

Step 1 — Recall $\cot \theta =\cos \theta /\sin \theta$. Why this step? $\cot$ measures how "horizontal" the ladder leans; as $\theta \to 90^{\circ }$ it shrinks to $0$, as $\theta \to 0^{\circ }$ it explodes to $\infty$ (cell C5).

Step 2 — Take the two limits. $\theta \to 90^{\circ }:\cot \theta \to 0⟹{\mu }_{\min }\to 0.$ $\theta \to 0^{\circ }:\cot \theta \to \infty ⟹{\mu }_{\min }\to \infty .$ Why this step? This tells us a vertical ladder barely needs friction; a floor-flat ladder needs impossible friction — it must slip. Matches everyday sense.

Step 3 — Plug in numbers. $\theta =60^{\circ }:{\mu }_{\min }=\frac{1}{2}\cot 60^{\circ }=\frac{1}{2}\cdot \frac{1}{\sqrt{3}}=\frac{1}{2\sqrt{3}}\approx \mathrm{0.289.}$ $\theta =30^{\circ }:{\mu }_{\min }=\frac{1}{2}\cot 30^{\circ }=\frac{1}{2}\cdot \sqrt{3}=\frac{\sqrt{3}}{2}\approx \mathrm{0.866.}$

Verify: The flatter ladder ($30^{\circ }$) needs $\approx 0.87$ vs the steeper one's $\approx 0.29$ — three times more friction. ✓ Consistent with the limits above.

---

Example 6 — A pure couple (cell C6)

Forecast: The two pushes are equal and opposite — so surely the wheel is balanced… right?

Look at the figure: the red circle is the wheel. The two black arrows are the equal-and-opposite pushes at the top and bottom. The dotted line between them is the separation $d=0.35\text{ m}$. Trace both arrows: they both try to spin the wheel the same way, so their turning effects add rather than cancel.

Step 1 — Check force balance. Why this step? $\sum \vec{F}=+8-8=0$. Translational equilibrium holds.

Step 2 — Compute the couple's torque. Why this step? A couple is two equal, opposite, parallel forces separated by a distance $d$ (see Couple and Moment of a Couple). Its torque is $Fd$ about any point — both forces turn the wheel the same way, so they add instead of cancelling (cell C6). Here $d=$ diameter $=0.35\text{ m}$. $\tau =Fd=8\times 0.35=2.8\text{N⋅m}.$

Step 3 — Conclusion. Net force $=0$ but net torque $=2.8\text{ N⋅m}\ne 0$ → NOT in equilibrium; the wheel angularly accelerates.

Verify: This is the parent's "steering wheel" intuition made numeric — the second condition is genuinely independent. ✓

---

Example 7 — Minimum friction as an inequality (cell C7)

Forecast: Bigger or smaller friction than a $45^{\circ }$ lean?

The four forces (name every symbol first).

• $W=200\text{ N}$: the beam's weight, down, at its centre.
• $N_f$: the floor normal force, up, at the foot.
• $N_w$: the wall normal force, horizontal (the wall is frictionless so it can only push straight out), at the top.
• $f$: the floor friction force, horizontal, at the foot, pointing toward the wall to stop the foot sliding out. This is the static-friction force from Static Friction; it can take any value up to a ceiling $\mu N_f$.

Step 1 — Vertical balance. Why this step? Only floor normal $N_f$ (up) and weight $W$ (down) are vertical: $N_f=W=200\text{ N}.$

Step 2 — Horizontal balance. Why this step? Only the friction $f$ (toward wall) and wall push $N_w$ (away from wall) are horizontal, so they must cancel: $f=N_w.$

Step 3 — Torque about the foot. Why this step? The foot carries $N_f$ and $f$ — both act there → moment arm $d=0$ → they vanish. Only $W$ and $N_w$ remain. Their moment arms: the weight acts at the beam's centre, whose horizontal distance from the foot is $\frac{L}{2}\cos \theta$; the wall push $N_w$ is horizontal and acts at the top, whose height above the foot is $L\sin \theta$. (These lengths carry the symbol $L$, the beam length.) $N_w(L\sin \theta )=W\frac{L}{2}\cos \theta .$

Step 4 — The length cancels; solve for $N_w$. Why this step? Every term has one factor of $L$, so it divides out — this is why the answer never needs a numerical length: $N_w=\frac{W}{2}\frac{\cos \theta }{\sin \theta }=\frac{W}{2}\cot \theta .$ With $\cot 50^{\circ }\approx 0.8391$: $N_w=\frac{200}{2}\times 0.8391=100\times 0.8391\approx 83.9\text{ N}.$

Step 5 — Apply the static-friction inequality to get ${\mu }_{\min }$. Why this step? Static friction can supply any force up to its ceiling $\mu N_f$, i.e. it must obey $f\le \mu N_f$ (see Static Friction). No-slip therefore requires $f\le \mu N_f⟹\mu \ge \frac{f}{N_f}.$ From Step 2, $f=N_w=\frac{W}{2}\cot \theta$, and from Step 1, $N_f=W$, so $\mu \ge \frac{\frac{W}{2}\cot \theta }{W}=\frac{1}{2}\cot \theta .$ Numerically, ${\mu }_{\min }=\frac{1}{2}\cot 50^{\circ }=0.5\times 0.8391\approx 0.42$.

Verify: ${\mu }_{\min }=\frac{1}{2}\cot 50^{\circ }=0.4196\approx 0.42$ ✓, and $N_w\approx 83.9\text{ N}$ ✓. At $45^{\circ }$ it would be $\frac{1}{2}\cot 45^{\circ }=0.5$; a steeper $50^{\circ }$ needs less ($0.42$). ✓

---

Example 8 — Real-world: the hanging sign (cell C8)

Forecast: Will $T$ be more or less than the total weight $120+180=300\text{ N}$?

Look at the figure: the hinge (black dot) is at the wall; the red arrow at the right end is the cable tension $T$, straight up because the cable is vertical. The two black downward arrows are the beam's weight (at the centre, $1.5\text{ m}$) and the sign's weight (at the right end, $3\text{ m}$). Since the cable is vertical and the beam horizontal, each moment arm is simply the along-the-beam distance from the hinge.

Step 1 — Translate words to forces. Why this step? Beam weight $120\text{ N}$ acts at the centre ($d=1.5\text{ m}$ from hinge); sign weight $180\text{ N}$ at the right end ($d=3\text{ m}$); cable tension $T$ upward at the right end ($d=3\text{ m}$); hinge reaction at the left ($d=0$).

Step 2 — Pivot at the hinge. Why this step? The hinge force acts at the hinge → $d=0$ → gone. Only $120$, $180$, and $T$ remain.

Step 3 — Torque balance (CCW positive). $T(3)-120(1.5)-180(3)=0.$ Why this step? $T$ lifts (CCW, +); both weights pull down (CW, −).

Step 4 — Solve. $3T=180+540=720⟹T=240\text{ N}.$

Verify: $T=240\text{ N}$ is less than the total weight $300\text{ N}$ — because the hinge shares the load. The hinge carries the remaining $300-240=60\text{ N}$ upward. Torque check: $240\times 3=720=120\times 1.5+180\times 3=180+540$. ✓

---

Example 9 — Exam twist: find BOTH support reactions (cell C9)

Forecast: Which support carries more — the one nearer the truck, or the far one?

Look at the figure: the two black triangles are supports $A$ (left) and $B$ (right), $10\text{ m}$ apart. The black up-arrows are the unknown reactions $R_A$ and $R_B$. The black down-arrow at the centre ($5\text{ m}$) is the beam's own weight; the red down-arrow at $3\text{ m}$ from $A$ is the truck. All forces are vertical and the beam horizontal, so every moment arm is just the along-the-beam distance from whichever pivot we pick.

Step 1 — Torque about $A$ to isolate $R_B$. Why this step? $R_A$ acts at $A$ → moment arm $d=0$ → it drops out (cell C2), leaving just one unknown, $R_B$. Beam weight acts at the centre ($d=5\text{ m}$), truck at $d=3\text{ m}$, and $R_B$ at $d=10\text{ m}$. Take counter-clockwise positive: $R_B(10)-500(5)-2000(3)=0.$ Why the signs? $R_B$ pushes up at the right end → turns CCW (+); both weights pull down → turn CW (−).

Step 2 — Solve for $R_B$. $10R_B=2500+6000=8500⟹R_B=850\text{ N}.$

Step 3 — Vertical force balance for $R_A$. Why this step? Translational equilibrium $\sum F_y=0$: total upward = total downward. $R_A+R_B=500+2000=2500⟹R_A=2500-850=1650\text{ N}.$

Step 4 — Cross-check by taking torque about $B$. Why this step? Because $\sum F=0$ already holds, torque is the same about any pivot (parent's "free pivot" theorem). Taking $B$ isolates $R_A$ and must give the same number: $R_A(10)=500(5)+2000(7)=2500+14000=16500⟹R_A=1650\text{ N}.$

Verify: Both routes give $R_A=1650\text{ N}$ ✓. And $R_A=1650>R_B=850$: the support nearer the truck carries more load, as expected. Total $1650+850=2500\text{ N}$ balances all the weight ✓.

---

Recall Did every cell get covered?

C1 signs — Ex 1,4 · C2 force through pivot — Ex 1,4,7,8,9 · C3 angled force — Ex 2 · C4 degenerate/zero — Ex 3 · C5 limiting — Ex 5 · C6 couple — Ex 6 · C7 friction inequality — Ex 7 · C8 word problem — Ex 8 · C9 both reactions — Ex 9. Every cell hit.