Rotational Equilibrium and Torque
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Direct answer
A rigid body is in complete equilibrium when two conditions hold simultaneously: the vector sum of forces is zero (ΣF = 0, translational equilibrium) and the vector sum of torques about any axis is zero (Στ = 0, rotational equilibrium). Torque itself is τ = r F sin θ, the product of force and the perpendicular distance (moment arm) between the axis and the line of action, measured in newton-metre. The liberating fact for NEET numericals is that the axis for Στ = 0 can be chosen anywhere — so a smart solver plants it on the line of action of the most inconvenient unknown force, which then drops out of the equation with zero moment arm. Beams on knife edges, seesaws, ladders against walls and hinged rods are the four stock settings, and all reduce to two or three torque equations.
What you must remember
- Two conditions: ΣF_x = 0, ΣF_y = 0 and Στ = 0 about any point; satisfying torque balance about one axis alone is not equilibrium.
- Torque formula: τ = r × F, magnitude rF sin θ; only the perpendicular component of force (or equivalently the perpendicular distance) produces turning effect.
- Axis freedom: the equilibrium equations hold for every choice of axis; strategically, pass the axis through unknown or unwanted forces (hinge, pivot) to eliminate them.
- Sign convention: fix one rotation sense (say anticlockwise positive) and stay with it — a weight acting to the left of the pivot and one to the right must enter with opposite signs.
- Centre of gravity shortcut: the entire weight of a uniform body acts at its geometric centre; for a non-uniform bar of mass M held by two supports, the torques of the two normal reactions about the centre of gravity must cancel.
- Couple: two equal, opposite, non-collinear forces produce zero net force but torque = F × (separation), which is why a couple causes pure rotation — opening a tap is the standard example.
- Ladder condition check: for a ladder of length L and mass m leaning at angle θ against a smooth wall, N_wall × L sin θ = mg × (L/2) cos θ, giving N_wall = mg/(2 tan θ); the floor then supplies friction f = N_wall horizontally.
A worked beam problem
A uniform 6 m beam of mass 20 kg rests on two supports, one at the left end A and the other 1.5 m from the right end B. A 30 kg load sits at the right end. Find both support reactions. Total downward force = (20 + 30)g = 50 × 9.8 = 490 N, so R_A + R_B = 490. Now take moments about A (so R_A vanishes): the beam's weight acts at the centre, 3 m from A, giving 20g × 3 = 588 N m clockwise; the load acts at 6 m, giving 30g × 6 = 1764 N m clockwise; R_B acts at 4.5 m anticlockwise. Balance: 4.5 R_B = 588 + 1764 = 2352, so R_B = 522.7 N — and immediately R_A = 490 − 522.7 = −32.7 N. The negative sign is the physics: support A must actually pull down (a bolt, not a mere rest), because the heavy load overhangs past R_B enough to tip the beam. This is the exam's quiet test — candidates who force both reactions positive get a "data inconsistency" feeling, while the torque method hands you the sign and the physical interpretation for free.
How the exam frames it
The recurring trap is the "smooth wall" phrase in ladder problems: smooth means zero friction at the wall, so the wall gives only a normal (horizontal) reaction, and the floor must supply all the balancing friction — students who add a vertical wall force break force balance silently. A second favourite is the point of application: torque of a force is independent of where along its line of action it is applied, so shifting a weight along its own vertical line changes nothing, an assertion-reason statement that appears almost verbatim. Watch also for the man-walking-on-a-plank genre: as he walks towards one support, that support's reaction grows linearly and the other's shrinks; the plank tips when the far reaction tries to go negative — the same sign logic as the beam above, dressed differently.
Frequently asked questions
What are the two conditions for a rigid body to be in equilibrium?
The net external force must be zero (ΣF = 0) and the net external torque about any axis must be zero (Στ = 0); either alone is insufficient.
Why can the axis for torque balance be chosen anywhere?
For a body in true equilibrium the torque sum vanishes about every axis; if it vanished about one axis only, a non-zero net force would still accelerate the body.
How does choosing the axis at a hinge simplify a problem?
Forces passing through the axis have zero moment arm (r = 0), so hinge reactions contribute no torque and the equation involves only the known weights and distances.
What is a couple and what is its torque?
A couple is a pair of equal, opposite forces with different lines of action; it produces zero net force but a pure torque F × d (d = separation of the lines of action), independent of the axis chosen.
Where does the weight of a uniform body act in torque equations?
At the geometric centre (centre of gravity); for a uniform rod or beam this is its midpoint, so the weight's moment arm is measured from the middle of the body.