5.3 Torque

Five little picture stories

Press Next (or Play) to walk through each story one small step at a time. The numbers come last.

1. Push the door at the handle

2. Push the door near the hinge

Quick check: you push a door at right angles, first 0.40 m from the hinge, then 0.80 m from the hinge, just as hard. The second push gives:

3. Push the door toward the hinge

4. The stuck wheel nut

Read the story as text

A wheel nut on a car is stuck. You pull on a short wrench with all your strength and nothing happens. Your neighbour slides a long pipe over the handle, pulls gently at the far end, and the nut turns.

Same idea with a heavy door: push next to the handle and it swings open. Push right next to the hinges, just as hard, and it hardly moves. Push along the door toward the hinge, and it does not turn at all.

What exactly decides how well a force makes something turn?

Quick check: a wrench is 0.20 m long. You pull at right angles on its end with 50 N. The torque is:

5. Check yourself

Think of your answer first, then tap to see it.

a) You push a door at right angles with the same force, once at the handle (0.80 m from the hinge) and once halfway (0.40 m). How do the torques compare?

Show answer

At the handle the torque is twice as big: τ = rF, and r is twice as big.

b) You push hard on the edge of a door, straight toward the hinge. What torque do you make?

Show answer

Zero. The line of your push goes through the hinge, so the lever arm is zero (θ = 0°, sin 0° = 0).

c) A wrench is 0.30 m long. You pull at right angles on its end with 50 N. What is the torque?

Show answer

τ = rF = (0.30 m)(50 N) = 15 N·m.

Already know this?

Three quick questions. Get all three right first time and you can skip ahead.

You push at right angles on a door with 20 N, 0.50 m from the hinge. The torque is:

You push hard on the edge of a door, straight toward the hinge. The torque is:

To get the same torque from a pull, you move your hand twice as far from the bolt. The pull you need is:

Torque is the turning effect of a force. It depends on how hard you push, where you push, and in which direction.

Force × lever arm

In pictures

  • Push the door at the handle, far from the hinge: it swings easily.
  • Push near the hinge just as hard: it hardly moves.
  • Push toward the hinge: it does not turn at all. Only the part of the push across the door turns it.
  • A long pipe on a wrench lets a gentle pull turn a stuck nut. Torque = how far out × how hard (the part across).

Below is the same idea in words, and then with numbers.

To turn something about an axis (a hinge, a bolt, a pivot), three things matter:

  1. How big the force is, F.
  2. How far from the axis it acts, r (measured from the axis to the point where the force is applied).
  3. The angle θ between r and F. Only the part of the force at right angles to r turns the object.

τ = r F sin θ

There are two handy ways to read this. Group F sin θ: it is F⊥, the part of the force perpendicular to r. Or group r sin θ: it is the lever arm r⊥, the shortest (perpendicular) distance from the axis to the line along which the force acts.

τ = r F⊥ = r⊥ F

θ r F lever arm r⊥ = r sin θ line of action

Sign: a torque that would turn the object counterclockwise is positive, clockwise is negative (state your choice). Units: newton-metres, N·m.

Special cases: θ = 90° gives the most torque (τ = rF). θ = 0° or 180° (pushing straight toward or away from the axis) gives zero torque. A force acting at the axis (r = 0) also gives zero torque.

With numbers: the wrench

A wrench is 0.25 m long (bolt to hand). You pull with 80 N.

Show all steps as text
  1. Pull at 90° to the handle: τ = rF sin 90° = (0.25)(80)(1) = 20 N·m.sin 90° = 1: all of the force turns the bolt.
  2. Pull at 60° to the handle: τ = (0.25)(80)(sin 60°) = (0.25)(80)(0.866) = 17 N·m.Only F sin 60° = 69 N is perpendicular to the handle.
  3. Same 20 N·m at 60°? F = τ / (r sin θ) = 20 / (0.25 × 0.866) = 92 N.A worse angle needs a bigger force.
  4. Pull along the handle (θ = 0°): τ = (0.25)(80)(0) = 0.The line of action passes through the bolt: lever arm zero.
  5. Use a 0.50 m pipe at 90° with only 40 N: τ = (0.50)(40) = 20 N·m, same as step 1 with half the force.Doubling r halves the force you need.

Push the swinging door

Top view of a swinging (café style) door, 1.0 m wide, 20 kg, hinge on the left. Its rotational inertia is I = ⅓ML² = 6.7 kg·m² (see 5.4). Choose where you push (r), how hard (F) and the angle θ between the door and your force (you keep that angle as the door turns). Counterclockwise (opening upward on screen) is positive. The sim stops when the door has swung 90°.

Try: θ = 90° then 30° (same r, F): torque drops by half. θ = 0° or 180°: no turn at all. θ = 270°: the door swings the other way (negative torque).

Worked examples

Opening a door basic

You push perpendicular to a door with 30 N at 0.80 m from the hinge. Find the torque.

Show solution
Show all steps as text
  1. τ = rF sin 90° = (0.80)(30) = 24 N·m.Perpendicular push, so sin θ = 1.

Bike pedal medium

A pedal crank is 0.17 m long. A rider presses straight down with 400 N. Find the torque when the crank makes 30° with the vertical, and the largest possible torque.

Show solution
Show all steps as text
  1. The force is vertical and the crank is 30° from vertical, so the angle between r and F is 30°.θ is between the crank (r) and the force, not the horizontal.
  2. τ = (0.17)(400)(sin 30°) = (0.17)(400)(0.50) = 34 N·m.Only half the push turns the crank.
  3. Largest torque when the crank is horizontal (θ = 90°): τ = (0.17)(400) = 68 N·m.That is why you push hardest with the pedal at the front.

Net torque on a metre stick medium

A metre stick pivots at its centre. Forces: 5.0 N down at 0.40 m left of the pivot; 8.0 N down at 0.20 m right; 3.0 N up at 0.30 m right. Find the net torque (counterclockwise positive).

Show solution
Show all steps as text
  1. 5.0 N down on the left turns it counterclockwise: +(0.40)(5.0) = +2.0 N·m.Picture the left end dropping: counterclockwise.
  2. 8.0 N down on the right turns it clockwise: −(0.20)(8.0) = −1.6 N·m.Right end dropping is clockwise.
  3. 3.0 N up on the right turns it counterclockwise: +(0.30)(3.0) = +0.90 N·m.Right end rising is counterclockwise.
  4. Στ = 2.0 − 1.6 + 0.90 = +1.3 N·m (counterclockwise).Add torques with their signs.

Beam held by a cable AP

A uniform 2.0 m beam weighing 50 N is hinged to a wall and held horizontal by a cable attached at its far end, pulling up and toward the wall at 30° above the horizontal. Find the cable tension using torques about the hinge.

Show solution
Show all steps as text
  1. Choose the hinge as the axis. The hinge force has r = 0, so it has no torque.Picking the axis where an unknown force acts removes it from the equation.
  2. Weight acts at the centre, 1.0 m out, perpendicular: τ = −(1.0)(50) = −50 N·m (clockwise).For a uniform beam, gravity acts at the middle.
  3. Cable: r = 2.0 m, angle between beam and cable = 30°: τ = +(2.0)(T)(sin 30°) = +1.0 T.Only T sin 30° is perpendicular to the beam.
  4. Not rotating, so Στ = 0: 1.0 T − 50 = 0, T = 50 N.More on this in 5.5. Note T is as big as the whole weight, even though it acts at twice the distance: at this shallow 30° angle only half of T (T sin 30°) is perpendicular to the beam.

Practice

  1. With the same size force, where and how should you push a door to get the biggest torque?

    Show answer

    τ = rF sin θ is biggest with large r and θ = 90°.

  2. A 10 N force acts 0.50 m from an axis at 30° to the line from the axis. The torque is:

    Show answer

    τ = (0.50)(10)(sin 30°) = 2.5 N·m. (4.3 N·m uses cos 30° by mistake.)

  3. A force's line of action passes straight through the axis of rotation. Its torque about that axis is:

    Show answer

    The lever arm (perpendicular distance from axis to line of action) is zero, so τ = r⊥F = 0.

  4. Force P acts at distance r, perpendicular. Force Q = 2P acts at distance r/2 at 30° to the radius. Compare the torques.

    Show answer

    τP = rP. τQ = (r/2)(2P)(sin 30°) = rP × 0.5. So τQ is half of τP.

  5. The "lever arm" of a force is:

    Show answer

    Lever arm r⊥ = r sin θ: the perpendicular distance from the axis to the line of action. The first choice is r itself, which is only the lever arm when θ = 90°.

  6. A seesaw pivots at its centre. A child pushes down on the right end. Taking counterclockwise as positive, the torque from this push is:

    Show answer

    Pushing the right end down turns the seesaw clockwise, which is negative with this sign choice.

  7. Short answer. Explain, using the torque equation, why a longer wrench makes a stuck bolt easier to turn.

    Show answer

    The bolt needs a certain torque to start turning. τ = rF sin θ, so for the same angle, a larger r gives the same torque with a smaller F. Doubling the length halves the force needed.

  8. Short answer. You need 45 N·m to loosen a nut. Your wrench is 0.30 m long and you can pull with at most 120 N. Can you do it? If not, what is the shortest wrench that works?

    Show answer

    Best case is θ = 90°: τmax = (0.30)(120) = 36 N·m < 45 N·m. No, it cannot be done.

    Shortest wrench: r = τ / F = 45 / 120 = 0.375 m (pulling at 90°).

AP question types for this topic

Four short free response questions, one of each AP Physics 1 type: Mathematical Routines, Translation Between Representations, Experimental Design and Analysis, and Qualitative/Quantitative Translation. Try each one on paper before you open the worked answer.

1. Mathematical Routines (5 points)

Nadia loosens a bolt with a wrench. She pulls on the end of the wrench, a distance L from the bolt, with a force F at an angle θ to the wrench handle.

  1. Derive an expression for the torque τ about the bolt in terms of F, L and θ.
  2. F = 40 N, L = 0.25 m and θ = 60°. Calculate the torque.
  3. Calculate the smallest force she could use at the same point to make the same torque, and say in which direction she should pull.
Show worked answer and scoring

(a) Only the part of F perpendicular to the handle turns the bolt: F⊥ = F sin θ. So τ = rF⊥ = FL sin θ.

(b) τ = 40 × 0.25 × sin 60° = 10 × 0.866 ≈ 8.7 N·m.

(c) The force is smallest when sin θ = 1, a pull at 90° to the handle: F = τ/L = 8.66/0.25 ≈ 35 N, perpendicular to the wrench.

Scoring (5 points): (a) 2 points: 1 point for using the perpendicular component, 1 point for τ = FL sin θ. (b) 1 point for 8.7 N·m. (c) 2 points: 1 point for perpendicular pull, 1 point for 35 N.

2. Translation Between Representations (5 points)

Tomas pulls on a door handle 0.80 m from the hinge with a steady 20 N force. He changes the angle θ between his pull and the door from 0° to 180°.

τθ
A
τθ
B
τθ
C
τθ
D

In words: Four graphs of torque against angle theta from 0 to 180 degrees. A is a straight line rising from the origin. B is a smooth hump: zero at both ends, highest in the middle. C is a flat line. D is a straight line falling to nearly zero.

  1. Which graph (A to D) shows the torque about the hinge against θ? Justify.
  2. At what angle is the torque largest, and what is its value?
  3. Calculate the torque at θ = 30°, and name the other angle that gives the same torque.
Show worked answer and scoring

(a) B. τ = rF sin θ: zero when he pulls along the door (0° or 180°), largest at 90°, and it follows the sine hump between. A and D never return to zero, and C ignores the angle.

(b) At 90°: τ = 0.80 × 20 = 16 N·m.

(c) τ = 16 × sin 30° = 8.0 N·m. sin 150° = sin 30°, so 150° gives the same torque (the hump is symmetric).

Scoring (5 points): (a) 2 points: 1 point for B, 1 point for τ = rF sin θ zero at the ends. (b) 1 point for 16 N·m at 90°. (c) 2 points: 1 point for 8.0 N·m, 1 point for 150°.

3. Experimental Design and Analysis (6 points)

Ayla balances a metre stick on a pivot at its centre. She hangs a 0.200 kg mass 0.30 m to the left of the pivot. She then pulls straight down with a spring scale at a distance r to the right of the pivot and records the force F that keeps the stick level.

r (m)0.100.150.200.300.40
F (N)5.93.93.01.951.5

In words: r of 0.10, 0.15, 0.20, 0.30 and 0.40 metres gave forces of 5.9, 3.9, 3.0, 1.95 and 1.5 newtons.

  1. Describe how she should measure r and F, and one way to reduce error.
  2. What should she graph to get a straight line? What does the slope equal?
  3. Use the data to find the slope, and compare it with the torque of the hanging mass.
Show worked answer and scoring

(a) Read r from the stick's scale to the scale's hook, measured from the pivot. Keep the scale vertical, wait for the stick to be level (a small spirit level helps), zero the scale before each reading, and repeat each reading and average.

(b) Level means Fr = τmass, so F = τ(1/r). Graph F against 1/r: a straight line through the origin with slope = τ of the hanging mass.

(c) 1/r = 10, 6.7, 5.0, 3.3, 2.5 m⁻¹. Slope ≈ (5.9 − 1.5)/(10 − 2.5) ≈ 0.59 N·m. The mass gives τ = 0.200 × 9.8 × 0.30 = 0.588 N·m: they agree within about 1%.

Scoring (6 points): (a) 2 points: 1 point for measuring r from the pivot with the stick level, 1 point for repeats or zeroing. (b) 2 points: 1 point for F against 1/r, 1 point for slope = τ. (c) 2 points: 1 point for the slope, 1 point for the comparison with 0.588 N·m.

4. Qualitative/Quantitative Translation (6 points)

A door's handle is 0.80 m from the hinge. Priya claims: “If I push at the middle of the door, 0.40 m from the hinge, I need twice the force to turn it the same way.”

  1. Explain in words why Priya is right. Assume she always pushes perpendicular to the door.
  2. Derive the force F needed for a torque τ at distance r, and find the forces at 0.80 m and 0.40 m for τ = 12 N·m.
  3. Find the force needed at 0.10 m from the hinge, and explain what happens to the force as she pushes closer and closer to the hinge.
Show worked answer and scoring

(a) Turning effect depends on both force and lever arm. Halving the lever arm halves the turning effect of the same force, so she needs double the force to make up for it.

(b) τ = rF so F = τ/r. At 0.80 m: 12/0.80 = 15 N. At 0.40 m: 12/0.40 = 30 N, exactly twice.

(c) 12/0.10 = 120 N. F = τ/r grows without limit as r → 0: pushing right at the hinge gives no torque however hard she pushes, the extreme case of her claim.

Scoring (6 points): (a) 2 points: 1 point for torque depending on lever arm, 1 point for half the arm needing double the force. (b) 2 points: 1 point for F = τ/r, 1 point for 15 N and 30 N. (c) 2 points: 1 point for 120 N, 1 point for the r → 0 link to the claim.

Common mistakes

The mistake: using sin θ with the wrong angle (for example the angle with the horizontal).

Why it is wrong: θ must be the angle between r (axis to point of application) and F.

How to spot it: sanity check: a force along r must give zero torque. If your formula gives the maximum there, swap sin and cos.

The mistake: forgetting the sign of each torque when adding.

Why it is wrong: torques that turn opposite ways cancel. Adding sizes overstates the net torque.

How to spot it: for every force, ask "if this were the only force, which way would it turn?"

The mistake: thinking a big force always gives a big torque.

Why it is wrong: a force through the axis, however big, gives zero torque (e.g. the hinge force on a door).

How to spot it: find the line of action and its perpendicular distance to the axis.

The mistake: calling torque "energy" because both use N·m.

Why it is wrong: in work, the force is along the displacement; in torque, the force is across the lever arm. Torque is never written in joules.

How to spot it: write torque as N·m and energy as J.