Rotational equilibrium

Some bodies start or stop turning more easily than others. Here the question is simpler: when does a body not turn at all? A body is in rotational equilibrium when the net torque about the pivot is zero, $\sum \tau = 0$.

This is Newton's first law for spinning. Zero net force keeps the center of mass moving steadily; zero net torque keeps the spin steady. "Steady" can mean at rest or turning at a constant rate, so equilibrium does not have to mean stopped.

Each force makes a torque equal to its size times its lever arm, the perpendicular distance from the pivot to the force's line of action, with a sign for which way it turns. Full balance needs two separate conditions: $\sum F = 0$ for straight-line motion and $\sum \tau = 0$ for turning.

To test rotational equilibrium, add up the torques. A torque is $\tau = F\,d$: force times its lever arm $d$, the perpendicular distance from the pivot to where the force acts. Two things must be right: every force that makes a torque is in the sum, and each one carries a sign.

Pick one direction as positive, say counterclockwise, and make clockwise negative. Then $\sum \tau = 0$ means the two directions cancel, not that every torque is zero. Torques that fight subtract; torques that help add. Do not forget the body's own weight: it acts at the center of mass and makes a torque whenever the pivot is not right under it.

The common slips are dropping a term, usually the beam's weight, flipping a sign so opposing torques add, or using the wrong distance. Each leaves the equation looking done but wrong.

$\sum \tau = 0$, with each torque $F\,d$ given a sign for its direction. Include the body's own weight, and let opposing torques subtract.

Full equilibrium is two separate statements at once: $\sum F = 0$ and $\sum \tau = 0$. Meeting one does not give you the other.

A body can have perfectly balanced forces and still spin. A couple is two equal, opposite forces at different points: $\sum F = 0$, but there is a real net torque, so the center of mass stays put while the body turns. A seesaw is not balanced by matching the children's weights either: a heavy child near the pivot loses to a lighter child far out, because torque is force times lever arm, not force alone.

Read $\sum \tau = 0$ correctly too: it means the spin rate is constant, which may be a steady turn, not a stop. Track straight-line motion and turning separately.

$\sum F = 0$ and $\sum \tau = 0$ are separate. Balanced forces can still leave a net torque, so balance turning with torques, not weights.

For a balanced body the net torque is zero about every point, not just one. Use that. Pivot right where an unknown force acts and that force gets a zero lever arm, so it drops out and leaves fewer unknowns.

So the support forces you find do not depend on which pivot you pick; they are real. A force acting at the pivot makes no torque, because its lever arm is zero. The common slip is measuring an arm from the wrong spot, a ruler's zero end or the object's middle, instead of from the pivot you chose.

Because the answer is the same for any valid pivot, a second pivot is a quick way to check your work.

$\sum \tau = 0$ about any point. Pivot where an unknown acts to drop it; a force at the pivot has zero arm; measure every arm from the pivot.

Take a uniform beam of weight $120$ N and length $4.0$ m, hinged at its left end and held up by a cable at its right end. The weight acts at the center, $2.0$ m from the hinge; the cable acts at $4.0$ m. About the hinge, $120(2.0) = T(4.0)$, so $T = 60$ N.

Now hang a $200$ N load $3.0$ m from the hinge. The torque equation becomes $120(2.0) + 200(3.0) = T(4.0)$, so $T(4.0) = 240 + 600 = 840$ and $T = 210$ N. A load far out costs more tension than its weight alone, because a longer arm means more torque.

Two traps to avoid: dropping the beam's own weight gives $T = 150$ N, too low; and measuring the load's arm from the cable instead of the hinge changes every term. Keep each arm measured from the hinge, and include every weight.

Sum torques about the hinge with each arm from the hinge: $120(2.0) + 200(3.0) = T(4.0)$ gives $T = 210$ N.

Balance problems break in ways that checking force alone will never reveal—each pitfall below carries a code the diagnostic targets.