C. Geometry and Algebra of Vectors · Dot and Cross Products

Torque

C2.6 C2.8About 65 minutes

Learning goals

I can explain torque as the cross product τ = r × F.
I can calculate the magnitude of a torque from |r|, |F| and θ, and the torque vector from components.
I can explain how to maximize torque, and what a ratchet does.
I can solve problems about wrenches, doors, pedals and balance.

Words to know

torque
The turning effect of a force about a pivot: τ = r × F, with magnitude |r||F| sin θ, in N·m.

pivot (axis of rotation)
The point or line an object turns about, such as a bolt or a door's hinges.

lever arm
The vector r from the pivot to the point where the force acts.

ratchet
A wrench mechanism that turns the bolt in one direction and swings back freely in the other.

Turning, not pushing

A force can make an object move, or it can make it turn. When you open a door, tighten a bolt or push a bicycle pedal, what matters is the turning effect, called torque (from the Latin torquere, to twist). You already know from experience that:

a longer wrench makes a stuck bolt easier to turn;
pushing a door near its hinges is much harder than pushing at the handle;
pushing along the handle of a wrench, toward the bolt, does not turn it at all.

The cross product captures all three facts in one formula.

Torque as a cross product

Let r be the vector from the pivot (the axis of rotation) to the point where a force F is applied. The torque is

τ = r × F

(τ is the Greek letter tau.) Its magnitude is

|τ| = |r||F| sin θ

where θ is the angle between r and F. Torque is measured in newton-metres (N·m). Torque is a vector: it points along the axis of rotation, with the direction given by the right-hand rule. If you curl the fingers of your right hand in the direction the object turns, your thumb points along τ.

The formula explains the three facts above: a longer wrench makes |r| bigger; pushing near the hinge makes |r| small; and pushing along the handle makes θ = 0° or 180°, so sin θ = 0.

Only the part of the force perpendicular to the wrench, |F| sin θ, turns the bolt. The torque is |r||F| sin θ, the magnitude of r × F.

Worked example

Example 1: A wrench

A mechanic pushes on the end of a 0.25 m wrench with a force of 80 N. Find the magnitude of the torque if the force is (a) perpendicular to the wrench and (b) at 60° to the wrench.

(a) |τ| = (0.25)(80) sin 90° = 20 N·m.
(b) |τ| = (0.25)(80) sin 60° = 20(√3/2) = 10√3 ≈ 17.3 N·m.

(a) 20 N·m, the largest possible with this force and wrench; (b) about 17.3 N·m.

Maximizing torque

To get the most torque from a given force: push as far from the pivot as you can (large |r|) and push perpendicular to r (θ = 90°, sin θ = 1). A ratchet wrench helps in tight spaces: it clicks back without turning the bolt, so you can make many short strokes, always pushing roughly perpendicular to the handle, without taking the wrench off.

Worked example

Example 2: Torque from components

A force F = [0, 40, −30] N is applied at the point with position r = [0.3, 0, 0] m from a pivot at the origin. Find the torque vector and its magnitude.

τ = r × F = [(0)(−30) − (0)(40), (0)(0) − (0.3)(−30), (0.3)(40) − (0)(0)] = [0, 9, 12].
|τ| = √(0 + 81 + 144) = √225 = 15.
Check with the angle: r · F = 0, so θ = 90°, and |r||F| sin θ = (0.3)(50)(1) = 15 ✓.

τ = [0, 9, 12] N·m, with magnitude 15 N·m. The object tends to turn about an axis in the direction [0, 9, 12].

Seen from above, a door turns about its hinge. A 25 N push at the handle, 0.8 m from the hinge, gives the same torque as a 100 N push only 0.2 m from the hinge. Arrow lengths are proportional to the forces.

Worked example

Example 3: Opening a door

A door handle is 0.8 m from the hinges. Pushing perpendicular to the door with 25 N at the handle opens it. What force would give the same torque if you pushed perpendicular to the door only 0.2 m from the hinges?

Torque at the handle: (0.8)(25) sin 90° = 20 N·m.
At 0.2 m: (0.2)|F| = 20, so |F| = 100 N.

100 N, four times the force at one quarter of the distance. That is why handles are placed far from the hinges.

Worked example

Example 4: The force needed

A wheel nut needs 54 N·m of torque to loosen. Using a 0.30 m wrench, you can only push at 75° to the handle. What force is needed?

54 = (0.30)|F| sin 75°.
|F| = 54 ÷ (0.30 sin 75°) ≈ 54 ÷ 0.2898 ≈ 186.3.

About 186 N. Pushing at 90° would need 54 ÷ 0.30 = 180 N, so the 15° difference costs only a few newtons, because sin 75° ≈ 0.966 is close to 1.

Worked example

Example 5: A bicycle pedal

A cyclist pushes straight down with 600 N on a pedal whose crank arm is 0.17 m long. Find the torque when the crank points (a) horizontally forward, (b) 30° above horizontal (forward), and (c) straight up.

(a) r is horizontal and F is vertical: θ = 90°, |τ| = (0.17)(600)(1) = 102 N·m.
(b) r points 30° above horizontal and F points straight down, so the angle between them is 30° + 90° = 120°. |τ| = 102 sin 120° ≈ 88.3 N·m.
(c) r points up and F points down: θ = 180°, sin 180° = 0, so |τ| = 0.

(a) 102 N·m, (b) about 88.3 N·m, (c) 0 N·m. This is why pushing down at the top of the stroke does nothing, and one reason some cyclists use clip-in pedals, which let them pull the pedal around the top and bottom of the stroke.

Just for fun

Torque and work are both measured in newton-metres, but they are different things. That's why engineers write torque as N·m and never as joules: calling it joules would really twist the meaning.

Hands-on activity

Door torque test

You need: an inside door that swings easily, a ruler or tape measure, and a strip of masking tape. A grown-up should be nearby. Keep fingers away from the hinge side and the gap where the door meets the frame.

Measure the distance from the hinge line to the handle. Put small pieces of tape on the door at the handle distance, at half of it and at a quarter of it.
With the door partly open, push it closed with one finger, perpendicular to the door, at each tape mark. Rank how hard each push felt.
Now push at the handle mark, but at a slant (about 45° to the door, as if pushing toward the hinge). Compare with the perpendicular push.
Use |τ| = |r||F| sin θ to explain each result. If the door needs about 5 N to swing when pushed at a 0.8 m handle, predict the force needed at each mark.

What to look for: pushing at half the distance needs about twice the force, and at a quarter about four times (10 N and 20 N in the example). The slanted push feels harder, because only |F| sin 45° ≈ 0.71|F| of it turns the door.

Your turn

Find the torque when a 50 N force is applied perpendicular to a 0.3 m wrench.

Find the torque when the same force is applied at 45° to the wrench.

A force F = [0, 0, 50] N acts at r = [0.2, 0, 0] m from a pivot. Find τ and |τ|.

A force is applied along the handle of a wrench, toward the bolt. What is the torque? Why?

A door is 0.9 m wide. You push perpendicular to it with 30 N at the outer edge. Find the torque about the hinges.

A bolt needs 40 N·m. With a 0.25 m wrench, what perpendicular force is needed?

More practice: warm-up

Write torque as a cross product.

What are the units of torque?

At what angle between r and F is the torque greatest?

What is the torque if F is parallel to r?

You double the length of a wrench and push with the same force at the same angle. What happens to the torque?

More practice: on your own

F = [0, 0, −20] N acts at r = [0.4, 0.3, 0] m. Find τ and |τ|.

A 0.35 m wrench is pushed at 70° to the handle. What force gives 60 N·m?

A child (weight 300 N) sits 1.5 m from the pivot of a seesaw. Where must a parent (weight 750 N) sit on the other side to balance it?

F = [3, 0, −1] N acts at r = [0.1, 0.2, 0.3] m. Find τ and |τ|.

A 90 N push on a 0.3 m wrench produces 20 N·m of torque. At what angle to the handle could the push be applied?

Challenge

Explain, using |τ| = |r||F| sin θ, why a mechanic might slip a pipe over a wrench handle to loosen a stuck bolt, and how a ratchet helps in a cramped engine.

Show that sliding a force along its own line of action does not change the torque.

Find the torque of F = [10, 0, 0] N about the origin if it acts at r = [0, 0.5, 0] m, and describe the axis of turning.

Check your understanding

Quiz

A 40 N force is applied perpendicular to a 0.5 m wrench. What is the magnitude of the torque?

How should you push on a wrench to get the most torque from a given force?

What is the torque when the force is parallel to r?

A force F = [0, 10, 0] N acts at r = [0.2, 0, 0] m. What is τ = r × F?

Which statement about torque is correct?

What does a ratchet wrench let you do?

For parents and teachers

Big idea (C2.6, C2.8): torque, the turning effect of a force, is the cross product τ = r × F, where r goes from the pivot to where the force acts. Its size is |r||F| sin θ in N·m, and it points along the axis of rotation by the right-hand rule.

How to help: The door activity takes five minutes and makes the formula unforgettable. In the kitchen, ask why a jar lid is easier to open with a wider grip, or why a long-handled tool helps. (A bigger |r| means more torque.)

Common difficulties:

Using the length of the force instead of the distance from the pivot.
Using cos θ instead of sin θ (torque uses the perpendicular part of the force).
Reversing the order: F × r points the opposite way.
Mixing up torque (N·m, a vector) with work (J, a scalar), even though the units look alike.