Torque and Rotational Equilibrium

Torque is the turning effect of a force about an axis, τ = rF sin θ = F × (perpendicular distance from the axis to the line of action of the force), measured in newton-meters (N·m), and a rigid body is in rotational equilibrium when the net torque about any axis is zero, in addition to the net force being zero. In brief: torque is largest when the force is perpendicular to the lever arm, and zero when its line of action passes through the axis. Counterclockwise torques are taken as positive and clockwise as negative. The two conditions for static equilibrium are ΣF = 0 and Στ = 0. The center of gravity can be treated as the point where the weight acts.

Opening a door, using a wrench, balancing a seesaw and designing a crane all involve torque. This guide defines torque and the lever arm, states the conditions of equilibrium, compares forces at different angles in a table, works through problems with a wrench and a loaded beam, and corrects common mistakes.

What is torque?

A force applied to a body can cause it to rotate about a pivot (axis). The effect depends on the size of the force, on the distance from the pivot, and on the direction. The torque is τ = rF sin θ, where r is the distance from the axis to the point where the force is applied, and θ is the angle between the position vector r and the force F. Equivalently, τ = F × d, where d = r sin θ is the lever arm (moment arm), the perpendicular distance from the axis to the line of action of the force. In vector form, τ = r × F, with the direction given by the right-hand rule along the axis. A longer wrench makes it easier to turn a bolt, and a door handle is placed far from the hinge. The unit is N·m (the same dimensions as energy, but different quantities: torque is not measured in joules).

What are the conditions for equilibrium?

An extended body is in static equilibrium when it does not accelerate linearly or angularly. Two conditions must hold:

1. Translational equilibrium: the vector sum of the forces is zero: ΣF_x = 0, ΣF_y = 0.
2. Rotational equilibrium: the sum of the torques about any point is zero: Στ = 0. If the sum of forces is zero, the choice of the axis is free, and a clever choice (for example a point where an unknown force acts) eliminates unknowns from the torque equation.

Examples: a seesaw balanced (clockwise torque equals anticlockwise torque), a beam supported at two points, a ladder leaning against a wall with friction at the foot, and a crane with a counterweight. A pair of equal and opposite forces with different lines of action is a couple, which produces pure torque (turning a steering wheel or tap with two hands) and no net force.

What is the role of the center of gravity?

For torque calculations, the weight of an extended body can be treated as a single force acting at its center of gravity (CG), which for a uniform body is its geometric center. A body supported at a point is balanced if its CG is vertically above or below the support. A body on a surface is stable (will not tip over) if the vertical line through the CG falls within the base of support. A lower CG and a wider base increase the stability; this is why racing cars are low and wide, and why a loaded truck is more likely to tip.

Torque at different angles

Worked example: wrench and beam

1. Wrench. A force of 80 N is applied perpendicular to a wrench at 0.25 m from the bolt: τ = 0.25 × 80 = 20 N·m. If the force is at 30° to the handle: τ = 0.25 × 80 × sin 30° = 10 N·m. With a longer wrench of 0.50 m and the same force: 40 N·m.
2. Seesaw. A child of mass 30 kg sits 2.0 m from the pivot. Where should a 40 kg child sit on the other side to balance? 30 × 2.0 = 40 × d, so d = 1.5 m.

3. Beam on two supports. A uniform beam of mass 20 kg and length 4.0 m rests on supports at its two ends. A 50 kg load is placed 1.0 m from the left end. Take torques about the left support to eliminate the unknown N₁: N₂ × 4.0 = (20 × 9.8) × 2.0 + (50 × 9.8) × 1.0 = 392 + 490 = 882, so N₂ = 220 N. Vertical forces: N₁ + N₂ = (20 + 50) × 9.8 = 686 N, so N₁ = 686 − 220 = 466 N.

4. Leaning ladder. A uniform ladder of mass 10 kg and length 5.0 m leans against a smooth wall at an angle of 60° to the ground. Torques about the foot: the wall force (horizontal) N_w acts at height 5.0 sin 60° = 4.33 m; the weight acts at a horizontal distance 2.5 cos 60° = 1.25 m. N_w × 4.33 = 98 × 1.25, so N_w = 28.3 N. Friction at the foot must supply f = N_w = 28.3 N, and the normal force at the floor is 98 N, so the coefficient of friction must be at least 28.3 ÷ 98 = 0.29.

5. Couple. Two equal and opposite forces of 5.0 N, 0.20 m apart, give a torque of 5.0 × 0.20 = 1.0 N·m.

How was torque developed?

Archimedes proved the law of the lever (about 250 BCE), which is the basis of the principle of moments: “Magnitudes are in equilibrium at distances reciprocally proportional to their weights.” The term “torque” (from the Latin torquere, to twist) was introduced in the 1880s by James Thomson, brother of Lord Kelvin. Leonardo da Vinci and Galileo analyzed beams and levers. Engineers use torque in the design of engines, transmissions and structures.

Which mistakes are common?

Mistake 1: Using the distance along the force instead of the perpendicular distance

The lever arm is perpendicular to the line of action of the force.

Mistake 2: Not choosing the axis wisely

Any axis works, but the one through an unknown force eliminates it.

Mistake 3: Forgetting the weight of the beam

The weight of a uniform beam acts at its center.

Mistake 4: Mixing the sign of torques

Clockwise and counterclockwise torques have opposite signs.