Use a force’s line of action instead of its full lever length

Choose the axis, reference or material model

Torque describes the tendency of a force to change rotation about a specified axis. For a lever on the positive x axis, a force at its end produces signed torque rFy. Equivalently τ=rFsinθ, where θ is the signed angle from the outward lever to the force. Counterclockwise torque is positive in this packet.

Build the algebraic relationship

The perpendicular moment arm is the shortest distance from the pivot to the line of action. It is not automatically the full lever radius. The same12 N force at r=0.5 m gives 3 N m at 30°,6 N m at 90° and zero at 0°. Reversing the sign of the angle reverses torque without changing the force magnitude.

Check the original numerical packet

A force along the lever can still require a pivot reaction even though it produces zero torque about that pivot. Likewise, zero net torque does not imply zero net force; translational and rotational balances are separate conditions. A pivot chosen as torque origin removes the pivot force’s moment but not the force from the recipient’s force balance.

Test the boundary and independent evidence

Torque uses N m and work can be expressed in the same dimensional units, but they are not interchangeable. Work requires displacement and is a scalar energy transfer; this workbench specifies only an instantaneous force geometry. To compute rotational work over a finite angle, one would need how torque changes along that motion, rather than multiplying this snapshot by an arbitrary angle without a model.

Worked original case

With r=0.75 m and force direction−30°, Fy=−6 N and signed torque−4.5 N m. The parallel force component along the lever does not add a turning effect.

Bounded investigation

Compare−90°,−30°,0°,30°,90° at one radius. Draw the force line and perpendicular pivot distance, then verify the same torque from rFy and F times signed moment arm. Keep all forces in the force balance even when their torque vanishes.

Changed-case numerical practice

Try the changed case before feedback

A 10 N force acts at radius 0.4 m at 30° to the outward lever. Find counterclockwise torque magnitude in N m.

Numerical result in the requested unitsCompare this result

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and limits

Original content checked 30 October 2026 against OpenStax College Physics 2e. All records are supplied or generated teaching evidence. No rotating equipment, applied loads, material tests or physical assembly is requested. An admissible virtual model state does not certify real equipment safety, supervised practical skill or independent mastery.

A separate rotation history: work and power

The snapshot above does not specify an angular displacement. For a separate ideal fixed-axis rotor, suppose a constant net torque of 2 N m acts through 3 rad in its positive direction. Predict the change in rotational kinetic energy before checking: net rotational work is torque times angular displacement, so it is 6 J. If the moment of inertia is 2 kg m² and the initial angular speed is 1 rad/s, the initial kinetic energy is 1 J and the final energy is 7 J; therefore the final speed is √7 rad/s. The instantaneous net power is torque times angular velocity, not torque alone: it starts at 2 W and ends at 2√7 W. These expressions assume a fixed axis and the stated constant torque; they do not assign work to the earlier single force snapshot without a path.

For a changed check, keep the same rotor and initial speed but apply 1 N m through 3 rad. The energy gain is 3 J, final kinetic energy is 4 J and final speed is 2 rad/s. Explain why doubling the work gain in the first packet did not double the final speed.