Unit 5 · Topic 5.3 Beta

Torque

Torque measures how well a force turns an object about an axis: τ = r⊥F = rF sin θ, where the lever arm r⊥ is the perpendicular distance from the axis to the force’s line of action.

Practice 1: Creating RepresentationsPractice 2: Mathematical RoutinesPractice 3: Scientific Questioning and Argumentation

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Question set for this topic

Part 1 · Hook

Why this matters

Try to open a heavy door by pushing right next to the hinges: it barely moves. Push the same amount at the handle and it swings. Push straight at the hinges and nothing happens at all. How hard you push is only half the story; where and in which direction you push decide whether anything turns.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. In a right triangle, the side opposite angle θ equals

  1. hypotenuse × sin θ
  2. hypotenuse × cos θ
  3. hypotenuse × tan θ
  4. hypotenuse ÷ sin θ
Show the answer

sin θ = opposite/hypotenuse.

  • Correct: hypotenuse × sin θ:
  • hypotenuse × cos θ:
  • hypotenuse × tan θ:
  • hypotenuse ÷ sin θ:

2. A free-body diagram shows

  1. each force on one object, drawn from a dot
  2. the net force as an extra arrow
  3. the velocity of the object
  4. forces the object exerts on others
Show the answer

An FBD shows only real forces on the object, from a dot.

  • Correct: each force on one object, drawn from a dot:
  • the net force as an extra arrow:
  • the velocity of the object:
  • forces the object exerts on others:

3. An object turns counterclockwise and slows down. Counterclockwise is positive. The sign of α is

  1. negative
  2. positive
  3. zero
  4. undefined
Show the answer

Slowing means α is opposite to ω.

  • Correct: negative:
  • positive:
  • zero:
  • undefined:

Part 4 · See it

See it first

A wrench grips a bolt, which is the axis of rotation. A hand pushes on the far end of the handle, a distance r from the axis, with a force F_A at angle θ = 60° to the handle, turning the wrench counterclockwise. A dashed line extends the force’s line of action. The lever arm, r⊥ = r sin θ, is the perpendicular distance from the axis to that line, drawn in amber with a right-angle mark. The panel gives τ = r⊥F = rF sin θ in N·m, notes that a counterclockwise torque is positive, and that a force whose line passes through the axis gives zero torque.
The lever arm is the perpendicular distance from the axis to the force’s line of action: r⊥ = r sin θ, and τ = r⊥F. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. A force acts on an object that can turn about an axisits line of action may pass the axis at some distance
  2. The line of action misses the axis by a perpendicular distance r⊥the force has a lever arm r⊥ = r sin θ
  3. The force has a lever armit exerts a torque τ = r⊥F that tends to turn the object one way
  4. Several forces acttheir torques add with signs (counterclockwise +, clockwise −) to a net torque
  5. A force’s line of action passes through the axisits lever arm and its torque are zero

Part 6 · Key ideas

Key ideas

  • Torque is the turning effect of a force: τ = r⊥F = rF sin θ, in N·m.
  • The lever arm r⊥ is the perpendicular distance from the axis to the line of action, not the distance to the point where the force acts.
  • Only the part of a force perpendicular to r turns the object.
  • Give each torque a sign: counterclockwise + and clockwise − (or the reverse, stated). Add them for the net torque.
  • A force diagram starts each force at its point of application, labeled with type and agent.

Part 7 · Misconception

A common mistake

The wrong idea: The torque of a force is its size times the distance from the axis to where it acts, whatever its direction.

What actually happens: Only the perpendicular distance from the axis to the force’s line of action counts: τ = rF sin θ. A force along the line toward the axis has no torque however far out it acts.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Diagram

Four ways to push a door

A door 0.80 m wide is hinged at its left edge. Four separate pushes, each 30 N, are shown; consider each on its own.

hinge (axis)21330°40.40 m0.80 m

1. What is the size of the torque from push 1 about the hinge? Give the value and its unit.

Type a number and its unit.

Show the answer

Push 1 is perpendicular to the door, so its lever arm is the full 0.80 m: τ = r⊥F = (0.80 m)(30 N) = 24 N·m.

  • Answer: 24 N·m

2. Rank the sizes of the torques the four pushes exert about the hinge.

  1. τ₁ > τ₂ = τ₃ > τ₄
  2. τ₁ = τ₃ > τ₂ > τ₄
  3. τ₁ = τ₂ = τ₃ = τ₄
  4. τ₄ > τ₁ > τ₃ > τ₂
Show the answer

Lever arms: push 1, 0.80 m; push 2, 0.40 m; push 3, (0.80 m)(sin 30°) = 0.40 m; push 4, 0. With F = 30 N: 24, 12, 12 and 0 N·m.

  • Correct: τ₁ > τ₂ = τ₃ > τ₄: Right: 24, 12, 12 and 0 N·m.
  • τ₁ = τ₃ > τ₂ > τ₄: Push 3 acts at the edge but at 30°, so its lever arm is halved to 0.40 m.
  • τ₁ = τ₂ = τ₃ = τ₄: Equal forces give equal torques only with equal lever arms.
  • τ₄ > τ₁ > τ₃ > τ₂: Push 4 points straight at the hinge, so its lever arm is zero.

3. Why does push 4 produce no torque about the hinge?

  1. Its line of action passes through the hinge
  2. It is weaker than the other three pushes
  3. It acts at the point farthest from the hinge
  4. The door’s weight cancels it
Show the answer

Torque is r⊥F, where r⊥ is the perpendicular distance from the axis to the force’s line of action. Push 4 points along the door straight at the hinge, so r⊥ = 0 and τ = 0, however hard the push.

  • Correct: Its line of action passes through the hinge: Right: the perpendicular distance from the hinge to its line is zero.
  • It is weaker than the other three pushes: All four pushes are 30 N; the difference is the lever arm.
  • It acts at the point farthest from the hinge: Pushes 1 and 3 act there too and do give torques.
  • The door’s weight cancels it: The door’s weight is vertical; push 4 is horizontal along the door. They do not cancel, and neither is the reason.

4. What is the size of the torque from push 3 about the hinge?

Type a number in N·m.

Show the answer

τ = rF sin θ = (0.80 m)(30 N)(sin 30°) = 12 N·m. Equivalently, the lever arm is (0.80 m)(sin 30°) = 0.40 m.

  • Answer: 12 N·m

5. The lever arm of a force about an axis is

  1. the perpendicular distance from the axis to the force’s line of action
  2. the distance from the axis to where the force acts
  3. the length of the force arrow
  4. the distance from the force to the center of mass
Show the answer

Extend the force along its line of action. The lever arm r⊥ is the perpendicular distance from the axis to that line; τ = r⊥F.

  • Correct: the perpendicular distance from the axis to the force’s line of action: Right: r⊥, the shortest distance from the axis to the line.
  • the distance from the axis to where the force acts: That is r. The lever arm is r⊥ = r sin θ, which equals r for a perpendicular force.
  • the length of the force arrow: The arrow’s length shows the force’s size, not a distance.
  • the distance from the force to the center of mass: Torque is about the chosen axis, not the center of mass.

6. How does a force diagram of a beam differ from a free-body diagram of the beam?

  1. Each force is drawn from the point where it acts on the beam
  2. Each force is drawn from a dot at the center of mass
  3. Each force is split into parts along and across the beam
  4. The net torque is drawn as an extra curved arrow
Show the answer

A free-body diagram puts all forces at one dot. For rotation the point of application matters, so a force diagram draws the object and starts each force where it acts, labeled with type and agent.

  • Correct: Each force is drawn from the point where it acts on the beam: Right: torques depend on where forces act, so the diagram shows it.
  • Each force is drawn from a dot at the center of mass: That is a free-body diagram; it hides where forces act, so it cannot show torques.
  • Each force is split into parts along and across the beam: Both diagrams show whole forces; components go beside the diagram.
  • The net torque is drawn as an extra curved arrow: Neither diagram shows a net force or net torque as an extra arrow.

7. Counterclockwise is positive. A force that would start a wheel turning clockwise exerts a torque that is

  1. negative
  2. positive
  3. zero
  4. positive or negative, depending on its size
Show the answer

On this exam a torque is described only as clockwise or counterclockwise. With counterclockwise positive, any torque tending to turn the object clockwise is negative.

  • Correct: negative: Right: the sign of a torque says which way it tends to turn the object.
  • positive: Positive was chosen as counterclockwise.
  • zero: A force that starts the wheel turning has a nonzero torque.
  • positive or negative, depending on its size: The sign comes from the direction of turning, not the size.

Part 9 · Summary

Summary

Torque measures how well a force turns an object about an axis: τ = r⊥F = rF sin θ, where the lever arm r⊥ is the perpendicular distance from the axis to the force’s line of action. Torques carry a sign for clockwise or counterclockwise and add to a net torque. A force diagram draws each force where it acts.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections