Visual walkthrough — Equilibrium of rigid bodies — translational + rotational

We assume you know only: a force is a push or a pull (an arrow), and objects have weight pulling them down. Everything else — torque, moment arm, couple, pivot-independence — we grow on the page.

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Step 1 — A force is an arrow, and it can do TWO different things

WHY we start here. Everyone's first instinct is "add up the arrows, if they cancel we're done." That instinct is only about effect (a) — sliding. It completely ignores effect (b) — spinning. The whole reason a rigid body needs two rules is that one arrow secretly controls two kinds of motion. We have to make effect (b) visible before anything else.

PICTURE. Below, the same blue arrow is drawn twice. On the left we track its sliding effect (where the door's centre goes). On the right we track its spinning effect (how much it turns about the red hinge).

Notice: the arrow's size governs the shove. But the spin depends on something the size alone doesn't tell you — how far from the hinge it acts, and at what slant. We need a number for that. Keep going.

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Step 2 — Building "turning power": the moment arm

WHY this number and not just "force $\times$ distance to pivot". Look at the picture: if you slide along the arrow's own line, the spinning effect never changes. So the spin can't depend on the full distance $r$ from pivot to where you happen to grab — it must depend only on the shortest (perpendicular) gap between the pivot and the line of action (the infinite line the arrow lives on). That perpendicular gap is the moment arm, written $d$.

PICTURE. The green segment is $d$. Watch how $d=r\sin \theta$ falls straight out of the right triangle: $\vec{r}$ (pivot → grab point) is the hypotenuse, $\theta$ is the angle between $\vec{r}$ and the force, and the side opposite $\theta$ is exactly the perpendicular drop $d$.

WHY $\sin \theta$ and not $\cos$ or the raw angle? We need a factor that is 1 when the force is perpendicular to $\vec{r}$ (best possible spin) and 0 when the force points straight along $\vec{r}$ (no spin — you're just pulling toward/away from the pivot). The one plain-trig function that is $1$ at $90^{\circ }$ and $0$ at $0^{\circ }$ is $\sin$. That is the entire reason it appears. See Torque for more.

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Step 3 — Two arrows that cancel as force but SPIN like crazy (the couple)

WHY this is the crucial step. This single picture is the proof that force balance alone is not equilibrium. It is the counterexample that forces a second rule into existence. This configuration has a name: a couple (see Couple and Moment of a Couple).

PICTURE. Top arrow points right, bottom points left, separated by $d$. As arrows: $+F$ and $-F$ sum to $0$. As spins: the top pushes the far side one way, the bottom pushes the near side the other way, and — this is the magic — those two pushes rotate the body in the same rotational sense. Their torques add.

So: a body can have $\sum \vec{F}=0$ and still be spinning up. Force balance is not enough. We are forced to demand a second, independent thing: torque balance.

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Step 4 — Where do these two rules actually come from? (Newton, promoted)

WHY sum over particles. A rigid body is millions of particles locked together. Internal forces between them come in equal-opposite pairs (Newton's 3rd law), so when we add all forces, the internal ones annihilate — only external forces survive.

PICTURE. The body as a cloud of dots; blue internal force pairs cancel arrow-to-arrow; only the outside (external) arrows remain.

Adding Newton's law $\vec{F}=m\vec{a}$ over all particles gives the whole-body statements:

$\underset{\text{total outside push}}{\underbrace{\sum {\vec{F}}_{\text{ext}}}}=\underset{\text{total mass}}{\underbrace{M}}\underset{\begin{array}{c}\text{accel. of the}\\ \text{centre of mass}\end{array}}{\underbrace{{\vec{a}}_{cm}}}\underset{\text{total outside twist}}{\underbrace{\sum {\vec{\tau }}_{\text{ext}}}}=\underset{\begin{array}{c}\text{moment of}\\ \text{inertia}\end{array}}{\underbrace{I}}\underset{\begin{array}{c}\text{angular}\\ \text{acceleration}\end{array}}{\underbrace{\vec{\alpha }}}$

The left equation is centre-of-mass motion; the right is its rotational twin, where $I$ is the Moment of Inertia. For the body to not start sliding we need ${\vec{a}}_{cm}=0$, and to not start spinning we need $\vec{\alpha }=0$:

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Step 5 — Choose your pivot freely (the beautiful fact)

WHY it works — the picture argument. Torque about a new point $B$ is torque about the old point $A$, plus a correction that is proportional to the total force. If the total force is zero, the correction vanishes — so the two answers must agree.

PICTURE. Point $A$ and point $B$, offset by the vector $\vec{d}$. Every particle's arm changes by the same shift $\vec{d}$. That common shift gets multiplied by $\sum {\vec{F}}_i=0$, so it disappears.

${\vec{\tau }}_B=\sum ({\vec{r}}_i-\vec{d})\times {\vec{F}}_i=\underset{={\vec{\tau }}_A}{\underbrace{\sum {\vec{r}}_i\times {\vec{F}}_i}}-\vec{d}\times \underset{=0}{\underbrace{\sum {\vec{F}}_i}}={\vec{\tau }}_A.$

Reading it: ${\vec{r}}_i$ = arm from $A$ to particle $i$; $\vec{d}$ = shift from $A$ to $B$; the second term dies because $\sum {\vec{F}}_i=0$. Conclusion: ${\vec{\tau }}_B={\vec{\tau }}_A$. Pivot choice is free. ∎

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Step 6 — Sign convention: turning it into arithmetic

WHY a sign at all. Two forces can twist in opposite senses; without signs they'd wrongly add up. Signs let opposite spins cancel — exactly what balance requires.

PICTURE. A green curved arrow (CCW, $+$) and a red curved arrow (CW, $-$) about the same pivot. For balance, the total CCW twisting must exactly equal the total CW twisting.

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Step 7 — Degenerate & edge cases (never get surprised)

We must cover the corner cases so no scenario ambushes you.

Case A — Force through the pivot ⇒ zero torque. If a force's line of action passes through the pivot, then $d=0$, so $\tau =F\cdot 0=0$. This is why the seesaw's support force and the ladder's floor reactions vanish when we pivot right at them.

Case B — Force parallel to $\vec{r}$ ($\theta =0$ or $180^{\circ }$) ⇒ zero torque. Here $\sin \theta =0$, so $\tau =Fr\sin \theta =0$. Pulling straight toward or away from the pivot never spins the body.

Case C — Force perpendicular ($\theta =90^{\circ }$) ⇒ maximum torque. $\sin 90^{\circ }=1$, so $\tau =Fr$, the biggest bang for the buck.

Case D — Balanced forces, unbalanced couple (the trap). $\sum \vec{F}=0$ but the body still spins (Step 3). Equilibrium fails. This is the case people forget.

PICTURE. Four mini-panels, one per case, with the moment arm $d$ and resulting $\tau$ labelled.

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The one-picture summary

Everything above, compressed: the two arrows of a couple (force cancels, twist survives) sitting beside the two boxed rules and the free-pivot fact.

Recall Feynman retelling — the whole walkthrough in plain words

Picture a floating door with a red hinge-dot. Step 1: one push does two jobs — it slides the whole door AND spins it about the dot. Step 2: the spin-power (torque) equals the push size times the perpendicular distance from the dot to the push's line — because sliding along the line changes nothing. Step 3: now push both edges in opposite directions equally: the door doesn't slide (pushes cancel) but it whirls — proof that "pushes cancel" is not the whole story. Step 4: promoting Newton's $F=ma$ to the whole body gives two twin laws — one for sliding, one for spinning — so "don't move" means both the total push AND the total twist must be zero. Step 5: and here's the gift — once the pushes already cancel, the total twist is the same no matter which dot you measure it about, so plant your dot right on the force you don't care about to make it vanish. Step 6: call anticlockwise twists $+$ and clockwise $-$ so opposites can cancel. That's the entire theory: forces freeze the sliding, torques tame the spinning.