Equilibrium of rigid bodies — translational + rotational

WHAT is equilibrium?

"Equilibrium" does not mean "at rest". It means zero acceleration — the body could be moving at constant velocity / spinning at constant rate. We focus on static equilibrium (at rest and staying at rest).

WHY two conditions? (Derivation from first principles)

Newton's second law for the whole body (sum over all particles, internal forces cancel by Newton's 3rd law):

$\sum \vec{F}_{\text{ext}} = M\vec{a}_{cm}$

For the centre of mass to have no acceleration, we need:

$\boxed{\sum \vec{F}_{\text{ext}} = 0}$

That fixes translation. But torque is what changes rotation. The rotational analogue of Newton's law is:

$\sum \vec{\tau}_{\text{ext}} = I\vec{\alpha}$

For no angular acceleration:

$\boxed{\sum \vec{\tau}_{\text{ext}} = 0}$

A beautiful fact: pivot choice is free

Proof. Torque about point $A$ vs about point $B$ (with $\vec r_B = \vec r_A - \vec d$, where $\vec d = \vec B - \vec A$): $\vec\tau_B = \sum (\vec r_i - \vec d)\times\vec F_i = \sum \vec r_i\times\vec F_i - \vec d\times\sum\vec F_i = \vec\tau_A - \vec d\times\underbrace{\sum\vec F_i}_{=0} = \vec\tau_A.$ So with translational equilibrium satisfied, net torque is independent of the chosen axis. ∎

Component form (the working equations)

For coplanar (2-D) forces in the $xy$-plane, the three scalar conditions are:

$\sum F_x = 0, \qquad \sum F_y = 0, \qquad \sum \tau_z = 0$

Sign convention: counter-clockwise torque = positive, clockwise = negative (consistent with $\hat z$ out of page).

Worked Example 1 — The seesaw (find the support distance)

A uniform plank, pivot at its centre. A 40 kg child sits 2 m left of the pivot. Where must a 25 kg child sit on the right to balance? ($g=10$)

Step 1 — Translational? The pivot supplies normal force $N$ upward; it auto-balances vertical forces, so $\sum F_y=0$ is satisfied once we know $N$. Why this step? We don't need $N$ for the question, so we won't use this equation.

Step 2 — Take torques about the pivot. Why? The pivot reaction $N$ passes through the pivot, so its moment arm is zero → it disappears. Smart axis choice!

$\sum\tau = 0:\quad (40)(10)(2) - (25)(10)(x) = 0$ Why this step? Left child turns one way (say +), right child the opposite way (−).

Step 3 — Solve. $800 = 250\,x \implies x = 3.2\text{ m}$ Why this answer is sensible: the lighter child sits farther out — larger moment arm compensates for smaller weight. ✓

Worked Example 2 — Leaning ladder (friction at the floor)

A uniform ladder of weight $W$, length $L$, leans at angle $\theta$ against a frictionless wall, on a rough floor. Find the minimum coefficient of friction $\mu$ so it does not slip.

Forces: weight $W$ at centre; wall normal $N_w$ (horizontal, since wall frictionless); floor normal $N_f$ (up); floor friction $f$ (horizontal, toward wall).

Step 1 — $\sum F_y=0$: $\;N_f = W$. Why? Only $N_f$ and $W$ are vertical.

Step 2 — $\sum F_x=0$: $\;f = N_w$. Why? Friction must hold back the wall's horizontal push.

Step 3 — Torque about the foot of the ladder. Why this axis? It eliminates $N_f$ and $f$ (both act at the foot → zero arm), leaving only $W$ and $N_w$.

Take counter-clockwise positive. Weight acts at horizontal distance $\tfrac{L}{2}\cos\theta$; wall normal acts at height $L\sin\theta$: $N_w\,(L\sin\theta) - W\left(\tfrac{L}{2}\cos\theta\right) = 0$ Why this step? Wall push tends to rotate the ladder up/over the foot; gravity tends to rotate it down.

Step 4 — Solve for $N_w$: $N_w = \frac{W\cos\theta}{2\sin\theta} = \frac{W}{2}\cot\theta.$

Step 5 — Combine: $f = N_w = \tfrac{W}{2}\cot\theta$ and $N_f = W$. No-slip needs $f \le \mu N_f$: $\mu \ge \frac{f}{N_f} = \frac{\tfrac{W}{2}\cot\theta}{W} = \frac{1}{2}\cot\theta.$ $\boxed{\mu_{\min} = \tfrac{1}{2}\cot\theta}$ Why sensible: steeper ladder (large $\theta$) → $\cot\theta$ small → less friction needed. A near-flat ladder needs huge $\mu$. ✓

Worked Example 3 — Couple = pure rotation

Two forces $F = 5$ N act on a rod, opposite directions, separated by $d = 0.3$ m.

$\sum\vec F = +F - F = 0$ ✓ (translational equilibrium of force), but $\sum\tau = F\cdot d = 5\times0.3 = 1.5$ N·m ≠ 0. Why this matters: the body is not in equilibrium — it angularly accelerates. This is the steel-man proof that you genuinely need the second condition.

Flashcards

Connections

• Torque — the rotational analogue of force; engine behind the 2nd condition.
• Centre of Mass — where weight effectively acts.
• Newton's Laws of Motion — $\sum F = Ma_{cm}$ underlies the 1st condition.
• Moment of Inertia — appears in $\sum\tau=I\alpha$; zero net τ ⇒ zero α.
• Couple and Moment of a Couple — counter-example showing why two conditions are needed.
• Static Friction — supplies the holding force in ladder/leaning problems.
• Centre of Gravity vs Centre of Mass — toppling vs sliding analysis.

Concept Map

Hinglish (regional understanding)