Question bank — Equilibrium of rigid bodies — translational + rotational

False — an extended body can still spin. A couple (two equal, opposite, offset forces) gives $\sum \vec{F}=0$ but a nonzero net twist $Fd$, so you also need $\sum \vec{\tau }=0$; see the couple figure above and Couple and Moment of a Couple.

False — it means zero acceleration, both linear and angular. A body gliding at constant velocity or spinning at a constant rate is in equilibrium; static equilibrium is the stricter case of being at rest and staying so.

False — a ball at the top of its throw is instantaneously at rest but has nonzero net force (gravity), so it is accelerating, not in equilibrium. Rest at an instant ≠ balanced forces.

True — the pivot-shift term is $\vec{d}\times \sum \vec{F}$, and with $\sum \vec{F}=0$ it vanishes (see the inline sketch above). So you may choose the cleverest axis.

False — only when $\sum \vec{F}=0$ is the torque axis-independent. With unbalanced forces the leftover $\vec{d}\times \sum \vec{F}$ makes torque genuinely change from point to point.

False — one force means $\sum \vec{F}\ne 0$, so translation is broken immediately, regardless of where it acts.

True for uniform bodies in uniform gravity — the centre of gravity coincides with the Centre of Mass there. In a non-uniform field they can differ; see Centre of Gravity vs Centre of Mass.

True — because a couple has $\sum \vec{F}=0$ by construction, its turning effect $Fd$ is independent of the axis you measure it about (same reason the pivot is free above).

True only if force balance already holds — then $\sum \vec{\tau }=0$ about one point forces it about all points. Without force balance, one axis is not enough.

False — a force whose line of action passes through the pivot has zero moment arm ($d=r\sin \theta =0$, since $\theta =0$), so it contributes no torque. This is why smart axis choice erases support reactions.

Error — torque is $\tau =rF\sin \theta =Fd$, using the perpendicular distance $d=r\sin \theta$, where $\theta$ is the angle between $\vec{r}$ and $\vec{F}$ (figure s01). Only the component of $\vec{F}$ perpendicular to $\vec{r}$ twists the body; see Torque.

Error — for a uniform ladder the weight acts at its Centre of Mass, i.e. its midpoint, giving moment arm $\frac{L}{2}\cos \theta$, not $L\cos \theta$ (that midpoint is exactly what produces the $\frac{1}{2}$).

Error — torque balance sets the two products weight×distance equal, not the weights. A light child far out balances a heavy child sitting close.

Error — if you take torques about the pivot, its reaction has zero moment arm and drops out entirely. Including it as a torque term wastes an unknown for nothing.

Error — constant angular velocity means $\vec{\alpha }=0$, so $\sum \vec{\tau }=0$; it is in rotational equilibrium. Motion is fine; only acceleration breaks equilibrium.

Error — the sign convention must be consistent across all torques in one equation. Fix one sense (say counter-clockwise $=+$) and stick to it throughout.

Error — a frictionless wall exerts only a normal (perpendicular) force. Friction lives at the rough floor, not the smooth wall.

Because a point has no size, so "where" a force acts is meaningless and it can only be pushed, never twisted. An extended body has geometry, so offset forces can spin it even when they cancel as pushes.

The support reaction acts through that point, giving it zero moment arm, so it vanishes from the torque equation. You solve for the quantity you want without ever introducing the reaction you don't.

Torque about the foot gives wall push $N_w=\frac{W}{2}\cot \theta$ — the $\frac{1}{2}$ comes from the weight acting at the ladder's midpoint (arm $\frac{L}{2}\cos \theta$) versus the wall's full-height arm $L\sin \theta$. Friction must match it ($f=N_w$), and $\cot \theta$ shrinks as $\theta \to 90^{\circ }$, so a steeper ladder needs less floor friction; see Static Friction.

If they act along different lines they form a couple: force cancels but each contributes twist in the same rotational sense, so torque adds to $Fd\ne 0$ (figure s02). Equilibrium demands both sums vanish.

Force changes linear motion of the Centre of Mass; the quantity that changes rotation is torque, which weights each force by its moment arm — see Newton's Laws of Motion and its rotational analogue $\sum \vec{\tau }=I\vec{\alpha }$.

By Newton's third law internal forces come in equal-opposite pairs that cancel in both the force sum and the torque sum, leaving only external forces to balance.

About that common point yes — every moment arm is zero. But you must still check force balance separately; concurrent forces can still fail to cancel.

They must be equal in magnitude, opposite in direction, and share the same line of action. Same line kills the couple; opposite-and-equal kills the net force.

They must be concurrent (pass through one common point), or be a special parallel set; otherwise a leftover couple appears and rotational balance fails.

No horizontal force can then oppose the wall's push, so $\sum F_x=0$ is impossible unless the wall push is zero, which needs the ladder vertical. Practically it slips; Static Friction at the floor is essential.

The minimum coefficient of static friction, ${\mu }_{\min }=\frac{1}{2}\cot \theta$ (the smallest ratio $f/N_f$ that prevents sliding), grows without bound as $\theta \to 0$, so no real surface can hold it — a flat leaning ladder must slip.

Yes — a couple's twist comes from the two applied forces, not from weight. With no gravity the body still angularly accelerates, since $\sum \vec{\tau }\ne 0$ regardless of where gravity acts.

Yes to both — a couple gives force balance without torque balance; a single off-centre force gives neither, and cleverly arranged forces can balance torque about a point while still sliding. The two conditions are independent.