Torque & Lever Arms

A force has a size and a direction; its torque has neither until you say what point it is taken about. Draw r from that point O to the place where the force acts and define τ = r×F. Take a 25 N force in the +y direction applied at (0.60 m, 0). About the origin, r = (0.60 m, 0) and τ_z = +15 N m. About A at (0.80 m, 0), r = (−0.20 m, 0) and τ_z = −5.0 N m: smaller, and turning the other way. About B at (0.60 m, −0.35 m), which lies on the force's own line, r is parallel to F and the torque is zero. One force, three answers, all of them correct. So every torque statement carries the words about O, and in a fixed-axis problem O sits on the axle, because that is the only rotation the body is free to make.

The cross product has magnitude r F sin φ, with φ the angle between r and F drawn tail to tail. Group it as r (F sin φ) and you have split the force: F cos φ points along r and merely pushes or pulls on the pivot, while F⊥ = F sin φ acts across r and does all the turning. Group it as (r sin φ) F and you have split the position instead: r⊥ = r sin φ is the perpendicular distance from O to the force's line of action, the lever arm. With r = 0.25 m, F = 80 N and φ = 50°, r⊥ = 0.192 m and F⊥ = 61.3 N, and both routes give 15.3 N m. Now slide the same force along its line until it acts 0.40 m from O. The angle falls to 28.6°, r sin φ is still 0.192 m, and the torque is still 15.3 N m. Only the line of action matters, and a line through O gives zero however large the force.

For forces in the xy-plane, r×F has only a z-component. Expanding (x î + y ĵ)×(Fₓ î + F_y ĵ) with î×ĵ = k̂ and ĵ×î = −k̂ gives τ_z = x F_y − y Fₓ, and with +z towards you positive means counterclockwise. Apply F = (30 N, −40 N) at (0.50 m, 0.20 m): τ_z = (0.50)(−40) − (0.20)(30) = −20 − 6 = −26 N m, clockwise. Check each piece by eye: a downward force to the right of O turns the body clockwise, and so does a rightward force above O. The magnitude route agrees, since r = 0.539 m, F = 50 N and φ = −74.9° measured from r round to F give r F sin φ = −26 N m. If you prefer to work with magnitudes, find r⊥ F first and then attach the sign by asking which way that force alone would turn the body about O.

Several forces on one body each get a torque about the same point, and the torques add with their signs. Take a 1.20 m horizontal boom hinged to a wall at O. A 150 N load hangs from the far end, the boom's 60 N weight acts at its middle, and a cable fixed 0.90 m out runs up and back to the wall at 35° to the boom; at this instant its tension is 400 N. The load contributes −(1.20)(150) = −180 N m and the weight −(0.60)(60) = −36 N m, both clockwise. The cable's lever arm is 0.90 sin 35° = 0.5162 m, so it contributes +206.5 N m. The hinge force, whatever it is, acts at O and contributes nothing. Net: −180 − 36 + 206.5 = −9.5 N m, clockwise: this tension is too small to keep the boom level, which would take 216/0.5162 = 418 N. Putting O at the hinge removed the one force nobody had told you, and that is the usual reason for choosing an axis.

Two forces of equal size and opposite direction, on parallel lines a perpendicular distance d apart, form a couple. Their sum is zero, so they cannot accelerate the centre of mass, yet about any point τ = r₁×F + r₂×(−F) = (r₁ − r₂)×F, and r₁ − r₂ joins the two points of application whatever O is. The magnitude is F d. Hands on a steering wheel pushing 12 N each way on lines 0.30 m apart give 3.6 N m about the hub, about the rim, or about a point a metre outside the wheel: there the lever arms are 1.30 m and 1.00 m, and 12 × 1.30 − 12 × 1.00 is still 3.6 N m. The general rule is τ′ = τ − R×Fₙₑₜ for a reference point moved by R. With zero net force the net torque belongs to the force system itself; with non-zero net force it changes as O moves, unless R happens to be parallel to Fₙₑₜ, so comparing torques taken about different points is meaningless.

Three limits on the single-arrow picture. A force spread along a body has no one point of application, so add the torque of each slice: τ = ∫ x dF. For gravity on a uniform 3.0 kg rod 1.20 m long, held horizontal from one end, dF = (M/L) g dx and τ = M g L/2 = 3.0 × 9.81 × 0.60 = 17.7 N m, exactly what one arrow M g at the centre of mass would give; for an uneven load the equivalent arrow sits wherever the integral puts it. Second, a hinge or axle permits rotation about its own line only, so only the torque component along that line counts. Push a door straight up, parallel to its hinges, and there is a torque about a point on the hinge but none along the hinge line, so the door does not swing. Third, N m here is not the joule. Torque is a cross product and a vector; work is a dot product and a scalar. They meet only when a torque acts through an angle, which is a later topic.