Mistake Master
Mistake Master
Two equilibrium conditions, not one. $\sum F = 0$ says the center of mass isn't accelerating. $\sum \tau = 0$ says the body isn't spinning up. They're independent: a body can satisfy one without the other. The cleanest case is a couple, two equal-and-opposite forces at offset points. The forces cancel, so $\sum F = 0$. But each one twists in the same rotational sense, so $\sum \tau \neq 0$. The center moves at constant velocity, staying put only if it started at rest, while the body spins.
Two questions to ask about any moving body. Is the center of mass accelerating? Is the body spinning up? The first is governed by $\sum F$; the second by $\sum \tau$. Each has its own equilibrium condition, and the two are independent.
Translational equilibrium: $\sum F = 0$, so $\vec a_\text{cm} = 0$. Rotational equilibrium: $\sum \tau = 0$, so $\alpha = 0$. A body can be in one without the other.
$\sum F = 0$ with $\sum \tau \neq 0$ gives a body whose center moves at constant velocity (zero, if it was initially at rest) while it spins up. $\sum \tau = 0$ with $\sum F \neq 0$ gives a body that translates without spinning. A static problem, where nothing moves, is just the case where both conditions hold at once.
Rotational equilibrium is Newton's first law for spinning:
$$\sum \tau = 0 \quad \iff \quad \omega \text{ is constant}.$$Zero net torque, no change in angular velocity. That covers $\omega = 0$ (a body at rest stays at rest) and constant nonzero $\omega$ (a wheel spinning at a steady rate has zero net torque on it).
When both $\sum F = 0$ and $\sum \tau = 0$ hold, $\sum \tau$ is the same (zero) about every axis. Use this: pick the axis that drops an unknown out of the equation, usually one that passes through an unknown force so its lever arm is zero.
A massless rod is pivoted at its middle. A $4$ kg mass hangs from the left side at $d_1 = 0.5$ m from the pivot. Where should a $2$ kg mass be placed on the right side so the rod is in rotational equilibrium?
$g = 10$ N/kg, CCW positive. The $4$ kg weight torques the rod CW with magnitude $m_1 g d_1 = (4)(10)(0.5) = 20$ N⋅m. The $2$ kg weight torques CCW with magnitude $m_2 g d_2 = 20\, d_2$. Set $\sum \tau = 0$:
$$-20 + 20\, d_2 = 0 \quad \Longrightarrow \quad d_2 = 1.0 \text{ m}.$$Lighter mass, longer arm. Halving the mass doubles the distance you need. Mass times arm is the balanced quantity, not mass alone.
A plank on two supports has two unknown reactions, $N_L$ and $N_R$, and only one $\sum F_y = 0$ equation. One equation, two unknowns: not solvable. The second equation comes from $\sum \tau = 0$. That's where rotational equilibrium earns its keep as a tool, not just a concept.
A uniform plank, mass $M = 20$ kg and length $L = 4.0$ m, rests on two supports: one at the left end ($x = 0$) and one at $x = 3.0$ m. A $10$ kg box sits on the plank at $x = 1.0$ m. Find the upward reaction $N_R$ at the right support.
Put the axis at the left support, so $N_L$ has zero lever arm and drops out. CCW positive; weights torque CW (negative).
$$\sum \tau = -Mg\!\left(\tfrac{L}{2}\right) - mg(1.0) + N_R(3.0) = 0.$$ $$N_R = \dfrac{(20)(10)(2.0) + (10)(10)(1.0)}{3.0} = \dfrac{500}{3} \approx 166.7 \text{ N}.$$Then $\sum F_y = 0$ gives $N_L = (M+m)g - N_R \approx 133.3$ N.
Right support carries more, $N_R \approx 167$ N against $N_L \approx 133$ N. It is not that both loads sit closer to the right: the box at $x = 1$ m is closer to the left, and only the plank's center at $x = 2$ m is closer to the right. The plank is the heavier load, so the two weights combine to act at $x = \tfrac{(200)(2) + (100)(1)}{300} \approx 1.67$ m, nearer the right support, and the support nearer the load carries more. Picking the axis at $N_L$ killed it from the torque equation, leaving one equation in one unknown. Then $\sum F_y = 0$ closes the system.
All three traps here come from forgetting that equilibrium is two conditions at once: balanced forces and balanced torques, with the lever arm carrying as much weight as the force.
$\sum F = 0$ doesn't imply $\sum \tau = 0$. A couple is the proof: two equal-and-opposite forces at offset points cancel as vectors but twist in the same rotational sense. $\sum F = 0$, $\sum \tau \neq 0$: the body's center doesn't accelerate, but it spins up.
Fix. Compute $\sum F$ and $\sum \tau$ as two separate checks. One doesn't tell you the other. If a problem asks about rotational equilibrium, the equation is $\sum \tau = 0$, full stop.
A shelf hinged to a wall and held by a cable has three unknowns: two hinge force components and the tension. $\sum F_x = 0$ and $\sum F_y = 0$ give two equations. The third has to come from $\sum \tau = 0$ or the system is underdetermined and the algebra stalls.
Fix. Count unknowns before you start. More than two unknown forces and you need $\sum \tau = 0$ too. The torque equation isn't optional decoration; it's the third equation in the system.
A $10$ N push at $0.10$ m from a hinge gives $\tau = 1.0$ N⋅m. A $2$ N push at $1.0$ m from the same hinge gives $\tau = 2.0$ N⋅m, double the first. Smaller force, longer arm, bigger torque. This is why door handles sit far from the hinge.
Fix. $\tau = r F \sin \theta$, or just $\tau = r F$ for a perpendicular force. Compare torques by comparing $rF$, not $F$ alone. Lever arm counts as much as force.