Answer: (b) and (c). A rigid body is in equilibrium only when both conditions hold: $\sum \vec{F}=0$ and $\sum \vec{\tau }=0$.

• Statement (a) alone is not enough — a body with zero net force can still spin up (a couple).
• Statement (b) states both conditions explicitly. ✓
• Statement (c) is the physical meaning of those two conditions: ${\vec{a}}_{cm}=0$ (from $\sum \vec{F}=0$) and $\vec{\alpha }=0$ (from $\sum \vec{\tau }=0$). ✓

Answer: (c). A push aimed straight at the hinge has its line of action passing through the pivot, so the moment arm $d=0$ and $\tau =F\cdot 0=0$. That is exactly why pushing a door toward its hinge never opens it. All other pushes have a nonzero perpendicular distance and so create torque.

Take torques about the pivot. Why this axis? The pivot's upward reaction force passes through the pivot, so its moment arm is zero and it drops out — no need to find it.

Left child spins the plank CCW (say $+$); right child spins it CW ($-$). Balance ($\sum \tau =0$): $(30)(10)(1.5)-(45)(10)(x)=0$ $450=450x⟹x=1.0\text{m}.$ Sanity check: the heavier child ($45$ kg) sits closer than the lighter child ($1.5$ m vs $1.0$ m). Heavier weight × smaller arm = lighter weight × bigger arm. ✓

Only the component of the force perpendicular to $\vec{r}$ twists the body. That perpendicular part has size $F\sin \theta$, so: $|\tau |=Fr\sin \theta =(20)(0.40)\sin 30^{\circ }=(20)(0.40)(0.5)=4.0\text{N⋅m}.$ Why $\sin$ and not $\cos$? The part of $\vec{F}$ along $\vec{r}$ (that's $F\cos \theta$) points straight at or away from the pivot — its line of action passes through the pivot, so it contributes nothing. Only the sideways part, $F\sin \theta$, has a moment arm.

The beam is uniform, so its weight $W=200\text{N}$ acts at the centre of mass, i.e. at the middle, $2.0\text{m}$ from the left. (See Centre of Gravity vs Centre of Mass.)

Step 1 — Torques about $A$ (left end). Why $A$? The reaction $N_A$ acts at $A$, so its moment arm is zero and it disappears — one unknown gone instantly. Only $N_B$ survives among the unknowns. Taking CCW positive: $N_B$ lifts the far end (CCW, $+$); the load and the weight pull down (CW, $-$): $N_B(4.0)-(300)(1.0)-(200)(2.0)=0$ $4N_B=300+400=700⟹N_B=175\text{N}.$

Step 2 — Translational balance $\sum F_y=0$: $N_A+N_B=300+200=500⟹N_A=500-175=325\text{N}.$

Cross-check with torques about $B$ (should give the same $N_A$, because $\sum F=0$ makes the axis choice free): $N_A(4.0)-(300)(3.0)-(200)(2.0)=0⟹4N_A=900+400=1300⟹N_A=325\text{N}.✓$ Sanity check: the load sits nearer to $A$, so $A$ carries more ($325>175$). ✓

Forces on the beam: weight $W=100\text{N}$ down at the centre ($L/2$); tension $T$ along the cable at the right end (angle $30^{\circ }$ above the beam); hinge force with unknown components $H_x$ (horizontal) and $H_y$ (vertical) at the left end.

Step 1 — Torques about the hinge. Why the hinge? Both hinge components act at the hinge, so they have zero moment arm and vanish — we solve for $T$ directly. The tension's vertical component $T\sin 30^{\circ }$ lifts the right end (CCW, $+$); weight pulls down at the middle (CW, $-$). The tension's horizontal component points straight along the beam toward the hinge, so it has zero moment arm about the hinge. $T\sin 30^{\circ }\cdot L-W\cdot \frac{L}{2}=0$ The $L$ cancels: $T\sin 30^{\circ }=\frac{W}{2}=50⟹T(0.5)=50⟹T=100\text{N}.$

Step 2 — Horizontal balance $\sum F_x=0$. The cable pulls the beam's right end toward the wall with horizontal size $T\cos 30^{\circ }$, so the hinge must push back: $H_x=T\cos 30^{\circ }=100\times \frac{\sqrt{3}}{2}\approx 86.6\text{N}.$

Step 3 — Vertical balance $\sum F_y=0$: $H_y+T\sin 30^{\circ }-W=0⟹H_y=100-100(0.5)=50\text{N (upward)}.$

Sanity check: cable and hinge together support the full $100\text{N}$ vertically: $T\sin 30^{\circ }+H_y=50+50=100$. ✓

Let the ladder have length $L$. Forces: weight $W$ at the middle; wall normal $N_w$ (horizontal, since the wall is frictionless); floor normal $N_f$ (up); floor friction $f$ (horizontal, pointing toward the wall to stop the foot sliding out).

(a) Vertical balance: only $N_f$ and $W$ are vertical: $N_f=W=200\text{N}.$ Torques about the foot (kills $N_f$ and $f$, both at the foot). The wall push acts at height $L\sin \theta$ and turns the ladder CCW over the foot ($+$); weight acts at horizontal distance $\frac{L}{2}\cos \theta$ and turns it CW ($-$): $N_w(L\sin \theta )-W\left(\frac{L}{2}\cos \theta \right)=0⟹N_w=\frac{W}{2}\cot \theta .$ With $\theta =60^{\circ }$, $\cot 60^{\circ }=1/\sqrt{3}\approx 0.5774$: $N_w=\frac{200}{2}(0.5774)=57.74\text{N}.$ Horizontal balance: $f=N_w=57.74\text{N}.$

(b) No-slip requires $f\le \mu N_f$, so the minimum is at equality: ${\mu }_{\min }=\frac{f}{N_f}=\frac{57.74}{200}=0.2887=\frac{1}{2}\cot 60^{\circ }.$ (See Static Friction for why $f\le \mu N$.)

(c) Now add a person $P=600\text{N}$ at distance $\frac{L}{4}$ up the ladder, i.e. at horizontal distance $\frac{L}{4}\cos \theta$ from the foot. Vertical: $N_f=W+P=200+600=800\text{N}.$ Torques about the foot: $N_w(L\sin \theta )=W\left(\frac{L}{2}\cos \theta \right)+P\left(\frac{L}{4}\cos \theta \right)$ $N_w=\cot \theta \left(\frac{W}{2}+\frac{P}{4}\right)=0.5774\left(100+150\right)=0.5774\times 250=144.34\text{N}.$ $f=N_w=144.34\text{N},{\mu }_{\min }=\frac{144.34}{800}=\mathrm{0.1804.}$

Surprising result: adding a person low on the ladder lowered ${\mu }_{\min }$ (from $0.289$ to $0.180$). Their weight boosts $N_f$ (denominator) a lot, while sitting near the foot adds little to the tipping torque. A person high up would do the opposite.

Limiting-case check: as $\theta \to 90^{\circ }$ (vertical ladder), $\cot \theta \to 0$, so ${\mu }_{\min }\to 0$ — a vertical ladder needs no friction. As $\theta \to 0^{\circ }$ (nearly flat), $\cot \theta \to \infty$, so ${\mu }_{\min }\to \infty$ — no real floor can hold a nearly flat leaning ladder. ✓