Statics & equilibrium
The idea
Bridges, cranes, ladders, and your own skeleton all obey the same two conditions: the forces sum to zero and the torques about any point sum to zero. The force condition you know from balanced-force problems; the torque condition is the genuinely new constraint, and it is what determines how a load is shared among supports. An object can have zero net force and still rotate — equilibrium demands both.
The strategic move is choosing where to take torques. Since the equilibrium conditions hold about every point, pick the pivot that makes unknown forces vanish: any force acting at the pivot has zero lever arm and drops out of the equation. Taking torques about a hinge eliminates both hinge force components at once, usually leaving a single unknown you can read off directly. For a uniform beam, gravity acts effectively at its center.
Students often worry that the answer depends on the pivot choice — it never does. Different pivots produce different-looking equations, but they are algebraically equivalent once the force equations are included. A good habit is to solve with one pivot and verify the torque balance about a second point.
Worked example
A uniform 4.0 m strut of mass 20 kg is attached to a wall by a hinge and held horizontal by a cable that runs from its far end to the wall, making a 30° angle with the strut. A 10 kg sign hangs from the far end. Using g = 9.8 m/s², find the cable tension and the hinge force components.
- Take torques about the hinge so both unknown hinge components drop out of the equation; only the cable and the two weights contribute.
- Balance the torques: the cable's perpendicular component T sin30° acts at 4.0 m, the strut's weight 20 × 9.8 = 196 N acts at its 2.0 m midpoint, and the 98 N sign acts at 4.0 m, so T sin30° × 4.0 = 196 × 2.0 + 98 × 4.0 = 392 + 392 = 784 N·m.
- Solve for tension: T × 0.5 × 4.0 = 2.0T = 784, giving T = 392 N.
- Use force balance for the hinge: horizontally, the hinge must cancel the cable's pull toward the wall, so H_x = T cos30° = 392 × 0.866 ≈ 339 N directed away from the wall.
- Vertically, the cable supplies T sin30° = 196 N of the 294 N total weight, so the hinge supplies H_y = 294 − 196 = 98 N upward.
- Check torques about the far end of the strut as an independent test: hinge vertical force 98 N at 4.0 m gives 392 N·m one way, strut weight 196 N at 2.0 m gives 392 N·m the other — balanced, so the solution is consistent.
Answer. The cable tension is 392 N, and the hinge pushes with about 339 N horizontally (away from the wall) and 98 N upward.
Check your understanding
- Why does taking torques about the hinge simplify the problem, and what would change if you chose the far end instead?
- How does the required cable tension behave as the cable angle approaches zero, and what does that tell a structural designer?
- Why is it legitimate to treat the strut's distributed weight as a single force at its midpoint?
- What additional condition decides whether a leaning ladder slips, beyond the two equilibrium equations?
Build the foundations first
Statics & equilibrium builds on these concepts. If any feel shaky, start there.