AP Physics C: Mechanics · Unit 5: Torque and Rotational Dynamics · Lesson 5.3

Deep Dive: Torque

🔬 Deep Dive
This is your textbook for this topic. Take your time. Read it more than once.
Back to Lesson 5.3
5.3.A.15.3.A.2Concept

Only the Perpendicular Force Creates Torque

Torque results only from the component of a force that's perpendicular to the position vector running from the axis of rotation to the point where the force is applied. Any component of the force that points along that same position vector — directly toward or away from the axis — contributes nothing to rotation at all.

This perpendicular distance has its own name: the lever arm — the perpendicular distance from the axis of rotation to the force's line of action.

Push far from the hinge — large lever armFlarge lever armPush near the hinge — small lever armFsmallSame force, wildly different torque — the lever arm is the whole story.
🔑This is exactly why pushing a door open near its handle takes so much less effort than pushing near the hinge — same force, but a dramatically larger lever arm far from the axis produces dramatically more torque.
5.3.B.15.3.B.1.i5.3.B.1.iiConcept

Force Diagrams for Rotational Systems

Force diagrams are the rotational cousin of the free-body diagrams you've drawn since Unit 2. They represent the relative magnitude and direction of every force acting on a rigid system — but they add one crucial extra piece of information: where each force is applied, relative to the axis of rotation.

💡This location detail is exactly what a standard free-body diagram leaves out, and exactly what matters most for torque — two identical forces can produce wildly different torques depending purely on where they're applied and at what angle.
5.3.B.25.3.B.2.i–iiiMath

The Cross Product: Magnitude and Direction

Formally, the torque exerted on a rigid system about a chosen pivot point is defined by the cross product between the position vector and the force:

τ⃗ = r⃗ × F⃗

Magnitude

The cross product between any two vectors A⃗ and B⃗ has a magnitude given by:

|A⃗ × B⃗| = AB sin θ

Applied to torque, this becomes τ = rF sin θ — exactly the lever-arm relationship from above, since r sin θ is precisely the lever arm.

Explore this directly below.

A force applied at the end of a position vector r, at an angle θ from r. Adjust r, F, and θ — watch torque peak at θ = 90° and vanish at θ = 0° or 180°, exactly matching this unit's header graph.

r0.60 m
F40 N
θ90°
rF
lever arm = r sin θ = 0.600 mτ = rF sin θ = 24.00 N·m

Maximum torque — the force is entirely perpendicular to r.

Direction

The cross product's direction is perpendicular to both vectors involved — normal to the plane that r⃗ and F⃗ define — and is found qualitatively using the right-hand rule: curl your right hand's fingers from r⃗ toward F⃗, and your thumb points in the direction of τ⃗.

Explore this directly below.

Curl your right hand's fingers from r toward F — your thumb points in the direction of τ. Adjust both vectors' directions and see the resulting rotation sense.

r direction0°
F direction90°
rFτ out of page (CCW)

On the AP exam, this direction is described simply as clockwise or counterclockwise about the axis — the full 3D right-hand-rule picture is useful for building intuition, but not required for the exam itself.

ExampleGuided Example — Calculating Torque

A wrench handle extends 0.25 m from a bolt. A worker applies a 60 N force at the end of the handle, at an angle of 40° above the handle's direction. Find the torque on the bolt.

Step 1 — Identify the relevant quantities
r = 0.25 m, F = 60 N, θ = 40° (angle between r and F)
ExampleWorked Example — Maximizing Torque

Using the same wrench and 60 N force from above, what angle should the worker push at to produce the maximum possible torque, and what is that maximum torque?

← Back to Lesson 5.3Ready for 5.4? Rotational Inertia introduces mass's rotational counterpart — how hard a system resists a change in its spin.