Static Equilibrium: Ladders, Beams & Hinges
- Apply the two conditions for static equilibrium to extended bodies
- Choose a pivot that eliminates unknown forces from the torque equation
- Solve ladder and beam problems including the required coefficient of friction
Two conditions, not one
A rigid body is in static equilibrium when it is neither translating nor rotating, which requires both ΣF = 0 and Στ = 0. Balanced forces alone are not enough: two equal and opposite forces applied at different points form a couple that produces rotation with no net force at all. Because forces have two components and torque one in a plane, a two-dimensional problem gives three independent equations and can therefore determine at most three unknowns.
Choosing the pivot is the whole trick
A force applied at the pivot has zero moment arm and contributes no torque. Since a body in equilibrium has zero net torque about any point, you may put the pivot wherever you like — so put it where an unknown force acts. A ladder problem with an unknown normal force and an unknown friction force at the base becomes a single equation in one unknown the moment you take torques about the base. Choosing well converts three simultaneous equations into three that solve one at a time.
Weight acts at the center of mass
For torque purposes the entire weight of a uniform object may be treated as acting at its center of mass, which for a uniform rod, ladder or beam is at its geometric center. This is what makes extended-body problems tractable: a continuous distribution of gravitational force is replaced by a single vector at a single point. For a non-uniform beam, or a beam with a load on it, treat each portion separately and give each its own moment arm.
A uniform 20 kg ladder 5.0 m long leans at 60° above the horizontal against a frictionless wall. Find the friction force at the base and the minimum coefficient of static friction that keeps it from slipping.
- 1.The wall is frictionless, so it exerts only a horizontal normal force N_w. Horizontally: f = N_w. Vertically: N_floor = mg = 20(9.8) = 196 N.
- 2.Take torques about the base, which eliminates both N_floor and f at once.
- 3.The wall force acts at height L sin 60° with moment arm L sin 60°; the weight acts at the midpoint with horizontal moment arm (L/2)cos 60°. So N_w L sin 60° = mg (L/2) cos 60°, and L cancels.
- 4.N_w = mg cos 60°/(2 sin 60°) = mg/(2 tan 60°) = 196/(2 × 1.732) ≈ 56.6 N, so f ≈ 56.6 N and μ_min = f/N_floor = 56.6/196 ≈ 0.29.
When solving a static equilibrium problem, the most useful place to put the pivot for the torque equation is —
Free-response equilibrium problems award points for the setup: a labeled free-body diagram, an explicit statement of the pivot chosen, and the torque equation with moment arms identified. Students who go straight to numbers routinely lose those points even when the final answer is right.
As a ladder leaning against a frictionless wall is made to stand more nearly vertical, the friction force required at its base —
Answer the 2 checkpoints as you read.
Sign in to save your progress