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A plank resting on two supports, a ladder leaning on a wall, a rod hinged at one end and held up by a string: these are the standard statics problems in Class 11 and JEE. Force balance alone cannot solve them. This post gives you a short method and works three problems with numbers you can check.
What are the conditions for equilibrium of a rigid body?
NCERT's chapter on systems of particles and rotational motion sets out two conditions for a rigid body in mechanical equilibrium:
Translational equilibrium: the vector sum of all forces is zero.
Rotational equilibrium: the vector sum of all torques is zero.
In a two-dimensional problem this gives three equations: forces along x, forces along y, and torques about any point you choose. The JEE (Main) 2026 syllabus lists equilibrium of rigid bodies in Unit 5, and the JEE (Advanced) 2026 syllabus lists it under mechanics. Check your exam year's bulletin.
Which point should you take torques about?
Any point works, because a body in equilibrium has zero net torque about every point. So choose the one where the most unknown forces act. A force through the pivot has zero moment arm and drops out of the torque equation.
Torque = force × perpendicular distance from the pivot to the line of the force. Take anticlockwise as positive and keep to it.
Worked example 1: a loaded beam
A uniform 4 m beam of mass 20 kg rests on supports at both ends. A 30 kg child sits 1 m from the left end. Take g = 10 m/s². Find the support forces.
Forces: beam weight 200 N at its centre (2 m from the left), child 300 N at 1 m, support forces L and R at the ends.
Torques about the left end, which removes L: R × 4 = 200 × 2 + 300 × 1 = 700, so R = 175 N.
Vertical forces: L + R = 500, so L = 325 N.
Check: the child sits nearer the left end, so the left support carries more.
Worked example 2: a ladder against a smooth wall
A uniform ladder 5 m long and weighing 200 N leans against a smooth wall with its foot 3 m from the wall, so its top is 4 m up. The floor is rough. Find the forces, and the least coefficient of friction that keeps it in place.
Forces: weight 200 N at the midpoint; wall force P, horizontal (a smooth wall can only push straight out); at the floor, normal force N up and friction f toward the wall.
Vertical: N = 200 N. Horizontal: f = P.
Torques about the foot, which removes N and f: P × 4 = 200 × 1.5, since the midpoint is 1.5 m from the wall horizontally. So P = 75 N and f = 75 N.
Least coefficient: μ = f/N = 75/200 = 0.375.
NCERT's own Example 6.9 solves a 3 m, 20 kg ladder against a frictionless wall in exactly this order. OpenStax University Physics has a similar ladder example on a slippery wall.
Now a 600 N person climbs. Standing 4 m along the ladder, the person is 4 × 3/5 = 2.4 m horizontally from the foot of the ladder. Torques about the foot: P × 4 = 300 + 600 × 2.4 = 1740, so P = 435 N. Now N = 800 N and f must be 435 N, so the friction needed rises to 435/800 ≈ 0.54.
If the floor's coefficient is only 0.4, the largest friction available is 0.4 × 800 = 320 N. Then 320 × 4 = 300 + 600 × 0.6d, where d is the distance climbed along the ladder, so d ≈ 2.7 m. Beyond that point the ladder slips. The higher you climb, the more friction the floor must supply.
Worked example 3: a hinged rod held by a string
A uniform 2 m rod of mass 10 kg is hinged to a wall and held horizontal by a string from its free end to a point on the wall above. The string makes 30° with the rod. Find the tension and the hinge force.
Forces: weight 100 N at 1 m from the hinge; tension T along the string; hinge force with unknown horizontal part H and vertical part V.
Torques about the hinge, which removes H and V: T sin 30° × 2 = 100 × 1, so T = 100 N.
Horizontal: the string pulls toward the wall with T cos 30° ≈ 86.6 N, so H ≈ 86.6 N away from the wall.
Vertical: V + T sin 30° = 100, so V = 50 N upward.
Hinge force magnitude: √(86.6² + 50²) = √(7500 + 2500) = 100 N.
The hinge force is neither along the rod nor purely vertical; the force equations give both parts.
What mistakes do students make in statics problems?
Using the distance along the rod instead of the perpendicular distance. On a slanted ladder, the moment arm of the weight is the horizontal distance, not half the length.
Giving a smooth wall a friction force. Smooth means the force is perpendicular to the surface only.
Assuming friction equals μN when the ladder is not about to slip. Static friction is only as large as needed, up to μN.
Putting the weight at the wrong point for a non-uniform body; find the centre of mass first.
Our post on common mistakes in rotational motion covers the torque errors that carry over to moving bodies.
A practice routine for equilibrium problems
Draw the body alone with every force, and mark where each one acts.
Circle the point where most unknowns act and take torques there first.
Solve three beam problems, three ladder problems and two hinged-rod problems.
Check each answer by taking torques about a second point; it should also give zero.
When you cannot see where to start, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer.
Today, take worked example 2 and find how far the person can climb if the floor's coefficient of friction is 0.5.
