AP Physics 1 · 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.
5.3.A.1Concept

What Is Torque?

Torque is the rotational analog of force. Just as a net force causes linear acceleration, a net torque causes angular acceleration. But torque is not just about how large the force is — it depends on where the force is applied and at what angle.

The classic example: a door. Push near the hinge and the door barely moves. Push at the outer edge with the same force and it swings easily. Push at the outer edge but parallel to the door (toward the hinge) and again nothing happens. Three different outcomes from the same force magnitude — because r and theta both change.

🔑Torque is measured in Newton-meters (N·m). Do not confuse this with Joules — both are N·m but torque is not energy. Torque is a rotational tendency; it becomes work only when the system actually rotates through an angle.
5.3.A.2Math

tau = rF sin theta

The magnitude of torque is:

tau = r * F * sin(theta)

Where r is the distance from the axis of rotation to the point where the force is applied, F is the force magnitude, and theta is the angle between the position vector r and the force vector F.

The sin(theta) factor extracts the component of F that is perpendicular to r — the only component that produces rotation. The component of F parallel to r (F cos theta) points directly toward or away from the axis and produces no torque.

Drag the sliders to see how r, F, and theta affect torque. Watch the perpendicular component (F_perp) and lever arm update live. Drag theta to 90° for maximum torque; to 0° for zero.

Radius r (m)0.8m
Force F (N)50.0N
Angle theta60.0°
F_perp = F sin(theta)
43.3 N
F_para = F cos(theta)
25.0 N
Lever arm = r sin(theta)
0.69 m
tau = r*F*sin(theta)
34.6 N·m
r=0.8mF=50N60°d⊥=0.69m

At theta=90° torque is maximum — F is fully perpendicular to r. At theta=0° or 180° torque is zero — F points along r, no rotation results.

ExampleWorked Example — Wrench on a Bolt

A 0.25 m wrench is used to tighten a bolt. A 80 N force is applied at the end of the wrench at 40° from the wrench handle. Calculate the torque.

5.3.A.3MathConcept

The Lever Arm

The lever arm (also called the moment arm) is the perpendicular distance from the axis of rotation to the line of action of the force — the line through which the force acts, extended in both directions.

tau = F * d_perp

This is mathematically identical to tau = rF sin(theta). The lever arm d_perp = r sin(theta). Substituting gives tau = F * (r sin theta) = rF sin theta. The two forms are the same equation viewed geometrically in two different ways.

💡When to use which form: Use tau = rF sin(theta) when you know the angle between r and F. Use tau = F * d_perp when you can directly identify the perpendicular distance from the axis to the line of action — this is often easier for horizontal forces with vertical distances and vice versa.
ExampleGuided Example — Lever Arm on a Seesaw

A 400 N child sits 1.5 m from the pivot of a seesaw. The weight force acts straight downward. What is the torque about the pivot?

Step 1Identify the lever arm
The force (weight) acts vertically downward. The axis is at the pivot. The perpendicular distance from the pivot to the line of action of the weight is the horizontal distance — 1.5 m. So d_perp = 1.5 m.
5.3.A.4Concept

Sign Convention and Net Torque

Torques have direction — they tend to produce either CCW or CW rotation. By convention:

CCW rotation tendency: tau = + positive
CW rotation tendency:  tau = − negative

When multiple torques act on a rigid system, the net torqueis their algebraic sum. This is what drives angular acceleration in 5.4: tau_net = I * alpha.

⚠️Always check which direction each torque would rotate the system about the specified axis before assigning its sign. The same force can produce a positive or negative torque depending on where the axis is and where the force is applied.
ExampleWorked Example — Net Torque on a Rod

A 2 m horizontal rod is pivoted at its left end. Force F1 = 30 N acts upward at the right end. Force F2 = 20 N acts downward at 0.8 m from the left end. Find the net torque about the pivot.

5.3.A.5Concept

Torque Force Diagrams

Torque force diagrams are like free-body diagrams, but with one critical additional requirement: forces must be drawn at their exact points of application, not at the center of mass. The location of the force on the object is what determines r — and r is what determines torque.

Standard FBD
Forces drawn at center of mass
Good for net force and linear acceleration
Location of application doesn't matter
Used in Units 2 and 4
Torque Force Diagram
Forces drawn at exact point of application
Required for torque calculations
Location on the object changes the torque
Used in Unit 5
🔑On AP FRQs, a torque force diagram question will explicitly tell you where the axis is and ask you to show all forces at their points of application. Draw the object, mark the axis, then draw each force arrow starting from exactly where it acts on the object.
← Back to Lesson 5.3Next: Lesson 5.4 →Rotational Inertia — resistance to angular acceleration.