An object at rest in equilibrium needs zero net force. That is not enough: two equal, opposite forces at the two ends of a ruler cancel, yet the ruler spins. So a second condition joins the first, written with torque.
Key idea
An object is in static equilibrium when both $\Sigma\vec{F} = 0$ (no acceleration) and $\Sigma\tau = 0$ (no angular acceleration). If the forces add to zero, the net torque is the same about every point, so you may take torques about whatever point is most convenient.
Why it works
Why can you pick any pivot? Move the reference point a distance $d$ along a beam. Each vertical force's torque changes by $-dF$, so the net torque changes by $-d\,\Sigma F$. When $\Sigma F = 0$, that change is zero. Smart choice: put the pivot where an unknown force acts. Its lever arm is then zero, it drops out of the torque equation, and you can solve for the other unknown at once.
- Draw every force on the object, including the force at the pivot or support and the object's own weight at its center of mass.
- Choose a pivot at an unknown force.
- Write $\Sigma\tau = 0$ (counterclockwise positive) and solve.
- Use $\Sigma F = 0$ for anything left, and as a check.
Worked examples
Worked example: A seesaw
A $30\ \text{kg}$ child sits $1.6\ \text{m}$ left of a seesaw's middle pivot. Where must a $40\ \text{kg}$ child sit to balance?
- The board's weight acts at the pivot, so it makes no torque there.
- $(30)(9.8)(1.6) = (40)(9.8)\,d$, so $d = \dfrac{30 \times 1.6}{40} = 1.2\ \text{m}$ to the right.
- The pivot pushes up with $(30 + 40 + m_{\text{board}})g$, from $\Sigma F = 0$.
Answer: $1.2\ \text{m}$ right of the pivot
Worked example: A plank on two supports
A uniform $20\ \text{kg}$ plank, $4.0\ \text{m}$ long, rests on supports at its ends. A $60\ \text{kg}$ painter stands $1.0\ \text{m}$ from the left end. Find both support forces.
- Pivot at the left support. Clockwise: the plank's weight $(20)(9.8) = 196\ \text{N}$ at $2.0\ \text{m}$ and the painter's $588\ \text{N}$ at $1.0\ \text{m}$. Counterclockwise: $F_R$ at $4.0\ \text{m}$.
- $$4.0F_R = (196)(2.0) + (588)(1.0) = 980,$$ so $F_R = 245\ \text{N}$.
- $\Sigma F = 0$: $F_L = 196 + 588 - 245 = 539\ \text{N}$. The support nearer the painter carries more.
Answer: $F_L = 539\ \text{N}$, $F_R = 245\ \text{N}$
Worked example: A sign on a hinged beam
A uniform $10\ \text{kg}$ beam, $2.0\ \text{m}$ long, is hinged to a wall and held level by a cable from its far end at $30^\circ$ to the beam. A $20\ \text{kg}$ sign hangs at the far end. Find the tension.
- Pivot at the hinge, so the unknown hinge force drops out.
- The cable's torque is $T(2.0)\sin 30^\circ = 1.0\,T$; the weights give $(98)(1.0) + (196)(2.0) = 490\ \text{N·m}$ the other way.
- $T = 490\ \text{N}$, more than the sign's weight, because the cable pulls at a shallow angle.
Answer: $T = 490\ \text{N}$
Common mistake
Thinking the choice of pivot changes the answer, or leaving out the force at the pivot. In equilibrium every pivot gives the same, correct answers; a clever pivot only makes the algebra shorter. But when you switch to a new point, the support force at the old pivot now has a lever arm and must be included. Check: solve once about each support; the two answers must satisfy $\Sigma F = 0$.
Balancing a stack of books on a moving cart is the same physics with motion added. Try the Book Cart in the Science Lab.