Static Equilibrium & Choosing a Pivot
- Apply both equilibrium conditions — zero net force and zero net torque — to a rigid body
- Choose a pivot point that eliminates an unknown force from the torque equation
- Solve beam, ladder and balance problems with distributed weight
Two conditions, not one
An object in static equilibrium is neither accelerating nor angularly accelerating, which means both ΣF = 0 and Στ = 0. Either can hold without the other. Two equal and opposite forces applied at different points on a rod give zero net force but a nonzero net torque — the rod spins in place without going anywhere. That configuration is called a couple, and it is why the second condition is a genuinely separate requirement rather than a consequence of the first.
Torque needs the perpendicular component
Torque is τ = rF sin θ, where θ is the angle between the position vector r and the force F. Only the component of the force perpendicular to r produces torque; a force pointed straight at or away from the pivot produces none, whatever its size. Equivalently, τ = F · r_⊥ where r_⊥ is the lever arm — the perpendicular distance from the pivot to the line of action of the force. The two formulations are the same statement, and the lever-arm version is usually faster to see on a diagram.
The pivot is yours to choose — choose it to erase an unknown
Because Στ = 0 about any point for a body in equilibrium, you may put the pivot wherever you like. Put it at the point of application of a force you do not know and do not want, and that force gets a lever arm of zero and disappears from the equation. A ladder problem with an unknown wall force and an unknown floor force becomes one equation in one unknown the moment you pivot at the floor. This is not a trick — it is the standard method, and it is what turns a two-unknown system into a one-line solve.
Distributed weight acts at the center of mass
A uniform beam's weight is spread along its whole length, but for torque purposes it behaves as if the entire weight acted at the center of mass — the midpoint for a uniform beam. For a non-uniform beam the center of mass is elsewhere, and locating it is often the first part of the question. Placing a beam's weight at the pivot or at the far end rather than at its center of mass is one of the most common errors in this unit.
A uniform 8.0 m beam of mass 20 kg rests on supports at each end. A 60 kg person stands 2.0 m from the left support. Find the upward force from each support. (g = 10 m/s²)
- 1.Pivot at the LEFT support so its force has zero lever arm and drops out.
- 2.Clockwise torques: person, (600 N)(2.0 m) = 1200 N·m; beam weight at the midpoint, (200 N)(4.0 m) = 800 N·m.
- 3.Counterclockwise torque: right support, F_R × 8.0 m.
- 4.Set equal: 8.0 F_R = 1200 + 800 = 2000, so F_R = 250 N.
- 5.Now use ΣF = 0: F_L + 250 = 600 + 200 = 800, so F_L = 550 N.
A force applied directly at the pivot never produces torque about that pivot, however large it is. That is the whole reason a door handle sits at the edge and not next to the hinge — and the reason pivoting at an unknown force is such an effective move.
A 20 N force is applied to a wrench at 30° to the handle, 0.40 m from the bolt. What torque does it produce?
When solving a beam problem with two unknown support forces, the best pivot choice is:
A 3.0 kg mass hangs 0.20 m from the fulcrum of a light balance beam. At what distance on the other side must a 2.0 kg mass hang to balance it?
State your pivot explicitly and label the sign convention (counterclockwise positive is standard). Rubrics award the correct torque equation, and an equation with an unstated pivot cannot be graded even when the numbers are right.
Answer the 3 checkpoints as you read.
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