Integrate signed torque across a distributed load

Choose the torque origin and direction

For a force on a point, torque about a specified origin is r cross F, with magnitude rF sin phi. In the plane, tau_z=xFy−yFx, positive according to the right-hand rule. A force component parallel to r produces no torque about that origin. Torque units N m match energy dimensions but torque is not work.

Build the differential load

Along a beam from 0 to 2 m, the signed vertical force density is f(x)=a+bx N/m. A small segment receives dFy=f dx; its endpoint torque is x f dx. Integrating yields total vertical force 2a+2b N and endpoint torque 2a+8b/3 N m, using the stated SI length scale.

Shift the origin without changing the forces

About the midpoint x=1, the lever arm is x−1. The torque becomes endpoint torque minus 1 m times the total force, hence 2b/3 N m. Torque generally depends on the reference point when net force is nonzero. Translation of the torque origin does not change which physical forces act.

Recognize a pure couple

With a=−1,b=1, total force is zero but torque is 2/3 N m. Upward and downward distributed contributions form a couple. Its torque is the same about all translated origins. Dividing torque by total force to assign an effective single-force location would divide by zero and is deliberately omitted; a finite single force cannot represent a pure couple.

Interpret a resultant only within its limits

When total force is nonzero, an effective vertical force at xeff=tau/F reproduces that force and torque in the plane. It may lie outside the physical loaded interval if positive and negative forces nearly cancel. It is a bookkeeping replacement, not evidence that the actual load acts at one point or that local bending stresses are equivalent.

Check the cross-product direction

The vector r cross F is perpendicular to both r and F. Curl the fingers of the right hand from r toward F through their smaller angle; the thumb gives the torque direction. For r=(2,0,0) m and F=(0,3,0) N, predict tau=(0,0,6) N m. Reversing only F reverses tau; a parallel force produces zero torque. The perpendicular distance from the axis to the force’s line of action is the lever arm: here it is 2 m. A recipient diagram must retain that application position as well as the force direction and magnitude.

Worked original case

a=1,b=1 gives total force 4 N, endpoint torque 14/3 N m and midpoint torque 2/3 N m. The equivalent force coordinate is 7/6 m.

Investigation

Compare a=1,b=1 with a=−1,b=1. Integrate force and torque about two origins; explain why the effective-force coordinate disappears for the second case. Retain signs of the load rather than converting all terms to magnitudes.

Public numerical transfer practice

Try a new practice case

A force(0,3) N acts at r=(2,1) m. What is torque z about the origin in N m?

Numerical result in the requested units Check this practice result

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Sources and evidence boundary

Original teaching and numerical cases checked 30 October 2026 against the official AP Physics C Mechanics framework and OpenStax fixed-axis rotation. Supplied/generated evidence is not physical measurement, a load rating or laboratory participation. No rotating machinery, hanging weights or mechanical construction is requested. Public feedback and retained self-reports do not certify independent learner mastery.