Introduction
Rigid body equilibrium requires both zero net force (no translation) and zero net torque (no rotation). These two vector conditions provide six scalar equations for 3D problems, determining equilibrium configurations. Understanding equilibrium is essential for structural engineering and stability analysis.
Conditions for Equilibrium
First condition: ΣF = 0 (translational equilibrium). Second condition: Στ = 0 about any point (rotational equilibrium). Six equations for 3D: three force components, three torque components. For 2D problems: two force equations, one torque equation.
Static Equilibrium
Object at rest remains at rest. Requires both force and torque balance. Stable equilibrium: small displacement produces restoring torque (returns to equilibrium). Unstable: small displacement produces torque away from equilibrium. Neutral: no torque for small displacement (indifferent equilibrium).
\nIntroduction
Rigid body equilibrium requires both zero net force (no translation) and zero net torque (no rotation). These two vector conditions provide six scalar equations for 3D problems, determining equilibrium configurations. Understanding equilibrium is essential for structural engineering and stability analysis.
Conditions for Equilibrium
First condition: ΣF = 0 (translational equilibrium). Second condition: Στ = 0 about any point (rotational equilibrium). Six equations for 3D: three force components, three torque components. For 2D problems: two force equations, one torque equation.
Static Equilibrium
Object at rest remains at rest. Requires both force and torque balance. Stable equilibrium: small displacement produces restoring torque (returns to equilibrium). Unstable: small displacement produces torque away from equilibrium. Neutral: no torque for small displacement (indifferent equilibrium).
\nChoosing Pivot Point
Torque balance can be calculated about any point. Strategic choice: pick point where unknown forces act to eliminate them from torque equation. Unknown forces through pivot have zero torque. This simplifies solving for unknown forces and torques.
Free Body Diagrams for Rigid Bodies
Show all external forces: gravity (at CM), normal forces, friction, tension, applied forces. Include force magnitudes and directions. Show torque arms. Identify unknowns. Write equilibrium equations systematically: force components first, then torque.
Examples
Ladder against wall: friction at floor, normal at wall and floor. Beam on supports: calculate support forces. Levers: mechanical advantage from torque balance. Mobiles: balance torques at each level. These are classic statics problems with many real-world applications.
\nSolved Example: Ladder Against Wall
A uniform ladder of mass m = 10 kg, length L = 5 m leans against a frictionless wall at angle θ = 60° with floor. Find: (a) Normal forces from wall and floor, (b) Friction force at floor, (c) Minimum coefficient of friction needed. Solution: Forces: N_w (wall, horizontal, no friction), N_f (floor, vertical), f (friction, horizontal), mg at center (down). Equilibrium: ΣF_x = f - N_w = 0 → f = N_w. ΣF_y = N_f - mg = 0 → N_f = mg = 98 N. Torque about bottom (eliminates N_f and f): N_w×Lsinθ - mg×(L/2)cosθ = 0. N_w×5×sin60° = 10×9.8×2.5×cos60°. N_w×4.33 = 245×0.5 = 122.5. N_w = 28.3 N. (a) N_w = 28.3 N, N_f = 98 N. (b) f = N_w = 28.3 N. (c) μ_min = f/N_f = 28.3/98 = 0.289. Ladder needs μ ≥ 0.289 to not slip.
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