# Torque and Equilibrium > Torque and equilibrium for JEE Physics: torque = r × F, the two equilibrium conditions, ladder problems and centre of gravity reasoning. - Canonical URL: https://prepelephant.com/topics/jee/physics/torque-and-equilibrium - Exam / course: JEE · Subject: Physics - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "Torque and Equilibrium", PrepElephant, https://prepelephant.com/topics/jee/physics/torque-and-equilibrium ## Direct answer Rotational equilibrium demands that the net torque about every axis vanish: tau = r F sin(theta), the turning effect of a force about a chosen axis, r × F with the lever arm built in. Complete equilibrium of a rigid body needs both conditions — sum of forces zero (translational) and sum of torques zero (rotational) — and the second may be checked about any axis you choose. Classic deployments: a uniform ladder of length L leaning at angle theta against a smooth wall with friction only on the floor stays put while tan(theta) is at least 1/(2 mu); and a couple — two equal opposite forces not in one line — produces pure rotation with zero net force. ## What you must remember - **Torque definition:** tau = r F sin(theta) = r × F; force through the axis (r and F parallel) exerts zero torque — the axis-placement weapon. - **Two conditions:** sum F = 0 and sum tau = 0 about any axis; choosing the axis at the intersection of two unknown forces eliminates both at once. - **Ladder, smooth wall:** with friction coefficient mu at the floor only, the wall reaction works out to N(wall) = (W/2) cot(theta) and the safety limit collapses to tan(theta) at least 1/(2 mu). - **Couple:** equal, opposite, non-collinear forces give zero net force but torque F d; a body under a couple purely rotates, and its torque is identical about every axis. - **Centre of gravity:** the point where the entire weight may be assumed to act; for equilibrium on a base, the vertical through the CG must fall inside the base — why a loaded truck tips before an empty one. - **Lami's theorem:** three non-parallel coplanar forces in equilibrium satisfy P/sin(alpha) = Q/sin(beta) = R/sin(gamma) — the one-line solution to many string problems. - **Torque and angular motion:** tau = dL/dt links this chapter to rotation; equilibrium is the special case of unchanging angular momentum. - **Pattern note:** Main asks seesaw and beam numericals; Advanced prefers ladders with friction at both surfaces and hinged rods held by strings. ## Choosing the axis wins the question A uniform rod of weight W and length L is hinged at one end to a wall and held horizontal by a string at the other end — find the string tension. Forces on the rod: W at the midpoint, tension T at the far end, and an unknown hinge reaction. Take torques about the hinge: the hinge force passes through the axis and vanishes from the equation, leaving T × L = W × (L/2), so T = W/2. The hinge reaction then follows from force balance if asked. The ladder is the same lesson with friction added. A uniform ladder against a smooth vertical wall, floor with coefficient mu, angle theta with the ground: the wall exerts only a normal force N(wall) horizontally. Torque balance about the foot: N(wall) L sin(theta) = W (L/2) cos(theta), so N(wall) = (W/2) cot(theta). Then horizontal force balance says floor friction f = N(wall), and safety demands f not exceeding mu N(floor) = mu W, which collapses to tan(theta) at least 1/(2 mu). ## Where students slip The lever arm is where errors concentrate: r sin(theta) is the perpendicular distance from the axis to the force's line of action, not the distance to the point of application. Second, each torque equation must commit to one axis only — mixing axes in one equation is meaningless. Third, the hinge or pivot reaction is not zero in general; it drops out only from the torque equation about its own hinge, not from force balance. ## Frequently asked questions ### What are the two conditions for a rigid body to be in complete equilibrium? The vector sum of all forces must be zero (no translation) and the sum of torques about any single axis must be zero (no rotation). ### Why can the torque equation be written about any arbitrary axis? Because when the net force is zero, the net torque is independent of the axis choice; strategically pick the axis through unknown forces to eliminate them. ### What is the slipping condition for a ladder against a smooth wall? With floor friction coefficient mu and ladder angle theta to the ground, equilibrium requires tan(theta) at least 1/(2 mu); below this the foot slides out. ### What is a couple in mechanics? A pair of equal and opposite forces whose lines of action do not coincide; the net force is zero but the torque F d is the same about every axis, producing pure rotation. ### When does a vehicle topple instead of skid? When the required friction exceeds mu times the normal reaction it skids, and when the vertical through the centre of gravity leaves the wheelbase it topples — whichever threshold is crossed first as the turn sharpens.