riaology
Unit 5 · Lesson 6 of 7

A perpendicular component creates torque.

A wrench is most effective when the force is perpendicular to its handle. Compare that with force along the handle.

Change the input. Follow the mathematics.

What you'll be able to do
  • Set up the model and calculate its torque magnitude.
  • Connect the equation to the changing visual.
  • Explain the assumptions and identify a misleading interpretation.
Predict before exploring

What will change if you increase angle between lever and force? Predict the direction, then explain the response.

A model you can changeMove one control at a time.
Vector components or geometric direction diagramReferencePerpendicular lever arm determines torque
Angle (degrees) versus Torque magnitude (N m). Torque magnitude. Selected torque magnitude: 10 N m. Torque uses the perpendicular lever arm. A force aimed along the lever has zero torque even when its magnitude is large.0059010180Torque magnitude (N m)Angle (degrees)
Torque magnitude

torque magnitude∣τ∣=0.5(20)sin⁡(90∘)=10 N m|\boldsymbol\tau|=0.5(20)\sin(90^\circ)=10\,\mathrm{N\,m}

Torque uses the perpendicular lever arm. A force aimed along the lever has zero torque even when its magnitude is large.

Assumptions, units, and limits

Lever length 0.5 m and force magnitude 20 N. Direction follows the right-hand rule, perpendicular to their plane.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

01

From the situation to an equation

The cross product's magnitude is the parallelogram area and its direction follows the right-hand rule. Applied to lever arm and force it gives torque. Reversing the factor order reverses direction; parallel factors give zero.

τ=r×F,∣τ∣=∣r∣∣F∣sin⁡θ\boldsymbol\tau=\mathbf r\times\mathbf F,\quad|\boldsymbol\tau|=|\mathbf r||\mathbf F|\sin\theta
02

Keep the reasoning visible

a×b=−b×a\mathbf a\times\mathbf b=-\mathbf b\times\mathbf a

Torque uses the perpendicular lever arm. A force aimed along the lever has zero torque even when its magnitude is large.

03

What makes this model useful?

Lever length 0.5 m and force magnitude 20 N. Direction follows the right-hand rule, perpendicular to their plane.

Cross-product order matters. Its magnitude is not the same as work, even though torque and energy share dimensional combinations.

A worked example

Calculate the starting scenario

  1. Use angle between lever and force = 90 degrees.
  2. ∣τ∣=0.5(20)sin⁡(90∘)=10 N m|\boldsymbol\tau|=0.5(20)\sin(90^\circ)=10\,\mathrm{N\,m}
  3. Torque uses the perpendicular lever arm. A force aimed along the lever has zero torque even when its magnitude is large. The checked quantity is torque magnitude.
Make the idea your own

Calculate. Compare. Explain.

Numeric checks allow 1% rounding error. Values near zero use a small absolute tolerance. Work through the setup before opening a solution.

Calculate and explain

Rebuild the starting calculation

Calculate the torque magnitude using angle between lever and force 90 degrees. State the meaning of its sign and units.

Checked quantity: torque magnitude · N m. Follow the definition for finite-cutoff and numerical-model questions.

Hint

The cross product's magnitude is the parallelogram area and its direction follows the right-hand rule. Applied to lever arm and force it gives torque. Reversing the factor order reverses direction; parallel factors give zero.

Explore the visual solution
Vector components or geometric direction diagramReferencePerpendicular lever arm determines torque
Angle (degrees) versus Torque magnitude (N m). Torque magnitude. Selected torque magnitude: 10 N m. Torque uses the perpendicular lever arm. A force aimed along the lever has zero torque even when its magnitude is large.0059010180Torque magnitude (N m)Angle (degrees)
Torque magnitude

torque magnitude∣τ∣=0.5(20)sin⁡(90∘)=10 N m|\boldsymbol\tau|=0.5(20)\sin(90^\circ)=10\,\mathrm{N\,m}

Torque uses the perpendicular lever arm. A force aimed along the lever has zero torque even when its magnitude is large.

Assumptions, units, and limits

Lever length 0.5 m and force magnitude 20 N. Direction follows the right-hand rule, perpendicular to their plane.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

The cross product's magnitude is the parallelogram area and its direction follows the right-hand rule. Applied to lever arm and force it gives torque. Reversing the factor order reverses direction; parallel factors give zero.

τ=r×F,∣τ∣=∣r∣∣F∣sin⁡θ\boldsymbol\tau=\mathbf r\times\mathbf F,\quad|\boldsymbol\tau|=|\mathbf r||\mathbf F|\sin\theta
Calculate and explain

Apply the idea to a new scenario

At angle 30° with 0.5 m lever and 20 N force, find torque magnitude.

Checked quantity: torque magnitude · N m. Follow the definition for finite-cutoff and numerical-model questions.

Hint

The cross product's magnitude is the parallelogram area and its direction follows the right-hand rule. Applied to lever arm and force it gives torque. Reversing the factor order reverses direction; parallel factors give zero.

Explore the visual solution
Vector components or geometric direction diagramReferencePerpendicular lever arm determines torque
Angle (degrees) versus Torque magnitude (N m). Torque magnitude. Selected torque magnitude: 5 N m. Torque uses the perpendicular lever arm. A force aimed along the lever has zero torque even when its magnitude is large.0059010180Torque magnitude (N m)Angle (degrees)
Torque magnitude

torque magnitude∣τ∣=0.5(20)sin⁡(30∘)=5 N m|\boldsymbol\tau|=0.5(20)\sin(30^\circ)=5\,\mathrm{N\,m}

Torque uses the perpendicular lever arm. A force aimed along the lever has zero torque even when its magnitude is large.

Assumptions, units, and limits

Lever length 0.5 m and force magnitude 20 N. Direction follows the right-hand rule, perpendicular to their plane.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

The cross product's magnitude is the parallelogram area and its direction follows the right-hand rule. Applied to lever arm and force it gives torque. Reversing the factor order reverses direction; parallel factors give zero.

τ=r×F,∣τ∣=∣r∣∣F∣sin⁡θ\boldsymbol\tau=\mathbf r\times\mathbf F,\quad|\boldsymbol\tau|=|\mathbf r||\mathbf F|\sin\theta
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Vector components or geometric direction diagramReferencePerpendicular lever arm determines torque
Angle (degrees) versus Torque magnitude (N m). Torque magnitude. Selected torque magnitude: 10 N m. Torque uses the perpendicular lever arm. A force aimed along the lever has zero torque even when its magnitude is large.0059010180Torque magnitude (N m)Angle (degrees)
Torque magnitude

torque magnitude∣τ∣=0.5(20)sin⁡(90∘)=10 N m|\boldsymbol\tau|=0.5(20)\sin(90^\circ)=10\,\mathrm{N\,m}

Torque uses the perpendicular lever arm. A force aimed along the lever has zero torque even when its magnitude is large.

Assumptions, units, and limits

Lever length 0.5 m and force magnitude 20 N. Direction follows the right-hand rule, perpendicular to their plane.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. The cross product's magnitude is the parallelogram area and its direction follows the right-hand rule. Applied to lever arm and force it gives torque. Reversing the factor order reverses direction; parallel factors give zero.

Your learning path

Can you explain the connection?

Mark complete when you can set up the calculation and explain what the visual is showing. Opening a page never completes it automatically.

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