Rotational Equilibrium set the two conditions for an object at rest: zero net force and zero net torque about any point. With Static Friction and the vector torques of Torque as a Cross Product, you can solve ladders, hinged beams and hanging signs. The Book Cart lab lets you feel what happens when the balance fails.
Key idea
For a rigid body in equilibrium: $\sum F_x = 0$, $\sum F_y = 0$ and $\sum\tau = 0$ about any point. Choose the pivot where the most unknown forces act: their torques vanish, often leaving one equation with one unknown.
Why it works
If the net force is zero, the net torque is the same about every point: moving the pivot by $\vec{a}$ changes each $\vec{r}$ to $\vec{r} - \vec{a}$, and the total torque changes by $-\vec{a}\times\sum\vec{F} = 0$. So you are free to pick the most convenient point. A force through the pivot has zero lever arm, so it drops out of the torque equation, which is how a smart choice removes unknowns you don't need.
Worked example: A ladder
A uniform $15\ \text{kg}$, $4.0\ \text{m}$ ladder leans against a frictionless wall at $60^\circ$ above the floor. Find the friction needed at the floor, and the least $\mu_s$.
- Forces: wall push $N_w$ (horizontal), floor normal $N_f$ (up), floor friction $f$ (toward the wall), weight $mg$ at the middle.
- Torques about the foot (removes $N_f$ and $f$): $N_wL\sin 60^\circ = mg\tfrac{L}{2}\cos 60^\circ$, so $N_w = \dfrac{mg}{2\tan 60^\circ} = \dfrac{147}{3.46} = 42.4\ \text{N}$.
- Horizontal: $f = N_w = 42.4\ \text{N}$. Vertical: $N_f = mg = 147\ \text{N}$.
- $\mu_s \ge 42.4/147 = 0.289$.
Answer: $f = 42.4\ \text{N}$; $\mu_s \ge 0.289$.
Worked example: A sign on a hinged beam
A uniform $20\ \text{kg}$, $3.0\ \text{m}$ beam sticks out from a wall on a hinge, held by a cable from its far end at $30^\circ$ to the beam. A $40\ \text{kg}$ sign hangs at the far end. Find the cable tension and the hinge force.
- Torques about the hinge: $T(3)\sin 30^\circ = 20(9.8)(1.5) + 40(9.8)(3)$, so $1.5T = 294 + 1176$ and $T = 980\ \text{N}$.
- Horizontal: the hinge pushes out with $H_x = T\cos 30^\circ = 849\ \text{N}$.
- Vertical: $H_y = 60(9.8) - T\sin 30^\circ = 588 - 490 = 98\ \text{N}$ (up).
- $|\vec{H}| = \sqrt{849^2 + 98^2} = 855\ \text{N}$, about $6.6^\circ$ above the beam's line.
Answer: $T = 980\ \text{N}$; hinge force $855\ \text{N}$.
Common mistake
Assuming the hinge's force points along the beam. That's only true for a light rod with forces at its two ends alone. A beam carrying its own weight and a load needs the hinge to push both outward and (usually) upward. Check: after finding $T$ from torques, write both force equations; if $H_y$ comes out nonzero, the hinge force is tilted off the beam's line.