Static Equilibrium & Torque Problems: JEE Main

Static equilibrium questions appear in JEE Main as standalone Mechanics problems and as components of angular momentum questions. The method is always the same: two conditions for equilibrium — net force equals zero and net torque about any point equals zero. The skill is choosing the right pivot to eliminate unknown forces and simplify the algebra. This guide teaches that skill with the problem types the exam actually uses.

Conditions for Static Equilibrium

For a rigid body to be in static equilibrium: (1) ΣF = 0 (vector sum of all external forces = 0); equivalently ΣFx = 0 and ΣFy = 0. (2) Στ = 0 about any chosen point (vector sum of all torques = 0). Torque: τ = r × F = rF sinθ where r is the position vector from the pivot to the force's point of application and θ is the angle between r and F. Sign convention: anticlockwise torque positive, clockwise negative (or choose consistently for each problem). The key insight: you may choose the pivot for condition (2) anywhere — always choose it at the point where an unknown force acts to eliminate that unknown from the torque equation.

Centre of gravity: the point at which the total weight can be considered to act. For a uniform body, it is the geometric centre. For a composite body, the CG is at r_cg = Σm_i r_i / Σm_i. JEE tests CG of L-shaped plates, hollow cylinders with one end closed, etc. — always treat as the sum of components and subtract removed parts. Take a free mechanics mock. For torque and angular momentum see our torque and angular momentum guide.

Standard Problem: Beam on Two Supports

A uniform beam of weight W and length L rests on two supports at distances d₁ and d₂ from the ends. Condition ΣFy = 0: R₁ + R₂ = W. Condition Στ = 0 about the left support: R₂ × (L − d₁ − d₂) = W × (L/2 − d₁). Solve for R₂, then R₁ = W − R₂. JEE often adds a load P at a specified point — treat it as an additional downward force and include its torque. The person-on-a-plank problem (plank on two spring scales, person walks) uses the same torque balance with changing load position.

Ladder Problem: A Classic

A uniform ladder of weight W and length L leans against a smooth wall at angle θ with the horizontal. Forces: weight W at centre (L/2 from foot), normal force from wall N_w (horizontal, at top), normal force from floor N_f (vertical, at foot), friction from floor f (horizontal, at foot). ΣFx = 0: f = N_w. ΣFy = 0: N_f = W. Στ = 0 about foot: N_w × L sinθ = W × (L/2) cosθ → N_w = W cosθ / (2 sinθ) = W/(2 tanθ). Friction coefficient required: μ ≥ f/N_f = N_w/W = 1/(2 tanθ). Maximum angle for stability: the ladder begins to slip when μ = 1/(2 tanθ_min) → tanθ_min = 1/(2μ). JEE has asked the exact angle at which a ladder begins to slip for a given coefficient of friction.

Hinge Reactions and Multiple Pivots

When a body is pinned at a hinge, the hinge can exert both horizontal and vertical force components — two unknowns. A system with a hinge, one support, and one applied load has three unknowns (hinge_x, hinge_y, reaction_at_support) and three equations (ΣFx, ΣFy, Στ). Take moments about the hinge to immediately eliminate hinge reactions from the torque equation. This is the most powerful move in static equilibrium — the pivot choice that removes two unknowns at once. For the full rotational mechanics chapter see our moment of inertia guide and our rotational motion guide.