Torque: push where it makes a difference
You will be able to: Calculate torque from force, application point and perpendicular lever arm.
Why is a door easier to turn near its handle?
Push a door near its hinges, then near its handle with the same sideways force. The farther push has a greater turning effect. Push directly toward the hinges and the turning effect disappears, even though the force can still be large.
A useful starting point: Resolving a force into components →
Words and symbols before equations
- Torque τ (tau)
- Turning effect of a force about a specified axis; unit N·m.
- Position vector r
- From the chosen axis to the force application point.
- Line of action
- The straight line through the force arrow, extended both ways.
- Lever arm r_perp
- Shortest perpendicular distance from the axis to the force’s line of action.
- Angle θ
- Angle between the outward position vector and the force, not an arbitrary angle to the surface.
What this picture assumes
Top view, pivot at left, force magnitude 10 N. r goes from pivot to application point. The dashed force line extends in both directions; the teal shortest segment is the perpendicular lever arm. Turning sense is counterclockwise for 0°<θ<180°.
Connect the picture to the physics
Only the component of force perpendicular to r produces a turning effect about the axis. Torque magnitude is |τ|=rF sinθ. Equivalently, |τ|=F r_perp, where r_perp=r sinθ. These are two ways to calculate the same torque, not two torques to add.
For a fixed r and F, a perpendicular push at θ=90° gives the greatest torque. At 0° or 180°, the line of action passes through the axis, r_perp=0, and torque is zero.
A force diagram for a rigid object must show where each force acts. Identify clockwise or counterclockwise tendency in the pictured plane and assign a sign for adding torques. We do not require a three-dimensional torque-vector direction. Although N·m has the same base units as a joule, torque is not energy; keep the torque unit N·m.
A worked example, step by step
A 10 N force acts 0.8 m from a pivot at 30° above the outward radius. Find its perpendicular component, lever arm and torque magnitude.
- F_perp=F sin30°=10(0.5)=5 N.
- Torque=rF_perp=(0.8)(5)=4 N·m.
- Alternatively, r_perp=r sin30°=0.4 m, so F r_perp=(10)(0.4)=4 N·m.
- From the top-view drawing, an upward component applied right of the pivot tends to turn counterclockwise.
The lever arm is perpendicular to the force line; it is not always the distance along the bar.
Can a nonzero force produce zero torque?
Compare with an explanation
Yes. Its line of action can pass through the chosen axis, making the perpendicular lever arm zero.
Predict. Change one thing. Explain.
Keep force at 10 N. Compare 0°, 30°, 90° and 180° at r=0.8 m. Trace the dashed line of action and teal lever arm; then move the force closer to the pivot.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
r=0.8 m; θ=90°. Lever arm=0.8 m, perpendicular force=10 N, torque magnitude=8 N·m. The force tends to turn the bar counterclockwise.
Top view, pivot at left, force magnitude 10 N. r goes from pivot to application point. The dashed force line extends in both directions; the teal shortest segment is the perpendicular lever arm. Turning sense is counterclockwise for 0°<θ<180°.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use an angular-motion or torque relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionAn 8 N force acts 0.5 m from a pivot at 30° to the outward radius. (a) Find F_perp. (b) Find r_perp. (c) Calculate torque both ways. (d) Explain the result if the same force points along the radius.
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Compare with the answer and four-point rubric
- 1 point: F_perp=4 N.
- 1 point: r_perp=0.25 m.
- 1 point: τ=(0.5)(4)=(8)(0.25)=2 N·m.
- 1 point: Torque is zero because the line of action passes through the axis.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Lever arm definition?
Perpendicular distance from axis to force line of action.
RECALL 2Which force component turns the object?
The component perpendicular to r.
RECALL 3Does torque depend on axis choice?
Yes; lever arms are measured from the selected axis.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Torque: push where it makes a difference
- |τ|=rF sinθ=F r_perp.
- Use the same axis for every torque you compare or add.
Remember: The lever arm is perpendicular to the force line; it is not always the distance along the bar.
Conditions: Top view, pivot at left, force magnitude 10 N. r goes from pivot to application point. The dashed force line extends in both directions; the teal shortest segment is the perpendicular lever arm. Turning sense is counterclockwise for 0°<θ<180°.
Refresh Kid · Unit 5 · Objectives 5.3.A, 5.3.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 5.3, objectives 5.3.A, 5.3.B. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Fall-2026 corrections also checked. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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