Define and Calculate Torque
- Understand torque as the rotational equivalent of force
- Apply $\tau = rF_\perp = rF\sin\theta$ correctly
Analyse Lever Arms and Angles
- Identify the perpendicular distance from pivot to line of action
- Determine when torque is maximised or zero
Solve Rotational Equilibrium Problems
- Apply the principle that net torque equals zero for balance
- Solve seesaw, door and wrench problems
Why is it easier to open a door by pushing near the handle than near the hinges? What quantity is different in each case?
Torque and Rotational Motion
Syllabus 5.22 — Investigate the relationship between the rotation of mechanical systems and the applied torque. Understand why force alone does not determine rotation.
What is Torque?
The rotational equivalent of force
Torque Diagram
Torque Detailed
A force can cause an object to move in a straight line (translation) or to spin (rotation). Torque is the quantity that causes rotation. It depends on three things:
- The magnitude of the applied force $F$
- The distance $r$ from the pivot (axis of rotation) to where the force is applied
- The angle $\theta$ between the force vector and the lever arm
Torque Formula
Key insight: Only the perpendicular component of force ($F_\perp = F\sin\theta$) contributes to torque. The component parallel to the lever arm ($F_\parallel = F\cos\theta$) pulls along the arm and does not cause rotation.
The HSC syllabus uses the notation $\tau = rF_\perp = rF\sin\theta$. You must be able to identify $r$, $F$ and $\theta$ in any diagram, and calculate torque for any angle. The unit of torque is the newton-metre (N m) — do not confuse this with joules, even though the units look the same. Torque is not energy.
Maximising and Minimising Torque
How angle and distance affect rotation
From $\tau = rF\sin\theta$, we can identify three special cases:
| Angle $\theta$ | $\sin\theta$ | Torque | Physical situation |
|---|---|---|---|
| $90°$ (force perpendicular to lever arm) | $1$ | $\tau_{\max} = rF$ | Pushing a door at right angles — maximum effectiveness |
| $0°$ or $180°$ (force parallel to lever arm) | $0$ | $\tau = 0$ | Pushing directly toward or away from the hinge — door does not rotate |
| $\theta$ between $0°$ and $90°$ | $0 < \sin\theta < 1$ | $0 < \tau < rF$ | Pushing a door at an angle — only the perpendicular component works |
Two ways to increase torque:
- Increase the force $F$ — push harder
- Increase the perpendicular distance $r$ — push farther from the pivot
Door handles are placed far from the hinges because this maximises $r$ and therefore maximises torque for the same pushing force. If you tried to open a door by pushing near the hinges, even a very large force would produce little rotation because $r$ is small.
Worked Example 1 — Wrench on a Nut
Calculating torque at different angles
Worked Example 2 — Seesaw in Equilibrium
Balancing torques for zero rotation
When an object is in rotational equilibrium, the sum of all torques acting on it is zero:
$\sum \tau = 0$
This means clockwise torques balance anticlockwise torques.
Connection to Circular Motion
Why torque matters for rotating systems
In circular motion, a net torque causes a change in angular velocity (angular acceleration). This is the rotational equivalent of Newton's second law:
Newton's Second Law for Rotation
Beyond syllabus: The moment of inertia $I$ depends on how mass is distributed relative to the axis. The syllabus focuses on the definition and calculation of torque ($\tau = rF\sin\theta$), not on rotational dynamics. However, understanding that torque causes rotation (just as force causes acceleration) helps connect this topic to uniform circular motion.
Centripetal force ($F = mv^2/r$) maintains circular motion by pulling toward the centre. It does not cause angular acceleration because it acts through the centre of rotation, producing zero torque ($r = 0$ for the centre). Torque is required only to start, stop or change the speed of rotation.
Common Misconceptions
Errors to avoid in torque problems
Misconception 1: "Torque is a force"
✗ Torque is not a force. It is the turning effect of a force. Two equal and opposite forces can produce zero net force but a large net torque (a couple).
Misconception 2: "The distance $r$ is along the force direction"
✗ the distance from the pivot to the point where the force is applied — measured along the lever arm, not along the force vector.
Misconception 3: "Torque is measured in joules"
✗ue has units of N m, it is not energy. The unit is called a newton-metre, not a joule. Torque and energy are fundamentally different quantities.
A force of 50 N is applied to a door handle 0.80 m from the hinges. Calculate the torque when the force is (a) perpendicular to the door, (b) at 60° to the door surface. Explain which pushing direction is most effective for opening the door.
A metre ruler is pivoted at the 50 cm mark. A 200 g mass is hung at the 20 cm mark. Where must a 150 g mass be placed to balance the ruler horizontally? Show all working and state the principle used.
Evaluate the statement: "A force always produces rotation if it is applied away from the pivot." Use the torque formula to identify a situation where a non-zero force applied away from the pivot produces zero torque. Explain the physical meaning of this result.
Model Answers
▾Question 1 (3 marks)
(a) $\tau = rF\sin\theta = 0.80 \times 50 \times \sin(90°) = 0.80 \times 50 \times 1 = \mathbf{40\ \text{N m}}$ (1 mark)
(b) $\tau = 0.80 \times 50 \times \sin(60°) = 0.80 \times 50 \times 0.866 = \mathbf{34.6\ \text{N m}}$ (1 mark)
The perpendicular push is most effective because $\sin(90°) = 1$ is the maximum value of $\sin\theta$. Any angle less than $90°$ reduces the perpendicular component and therefore reduces torque (1 mark).
Question 2 (4 marks)
Principle: For rotational equilibrium, the sum of clockwise torques equals the sum of anticlockwise torques ($\sum \tau = 0$) (1 mark).
Distance of 200 g mass from pivot: $r_1 = 50 - 20 = 30\ \text{cm} = 0.30\ \text{m}$
Let $r_2$ be the distance of the 150 g mass on the opposite side.
$r_1 m_1 g = r_2 m_2 g \Rightarrow r_2 = \dfrac{r_1 m_1}{m_2} = \dfrac{0.30 \times 0.200}{0.150} = \dfrac{0.060}{0.150} = \mathbf{0.40\ \text{m}}$ (2 marks for method + answer)
Position: $50 + 40 = \mathbf{90\ \text{cm}}$ mark (or 40 cm to the right of the pivot) (1 mark).
Question 3 (5 marks)
The statement is incorrect (1 mark). From $\tau = rF\sin\theta$, torque depends on the angle $\theta$ between the force and the lever arm, not just on the magnitude of the force or its distance from the pivot (1 mark).
When $\theta = 0°$ or $180°$ (force directed along the lever arm, either toward or away from the pivot), $\sin\theta = 0$ and therefore $\tau = 0$ (1 mark).
Physical meaning: A force directed through or away from the pivot pulls along the lever arm but does not tend to rotate the object (1 mark). For example, pushing a door directly toward its hinges produces no rotation regardless of how hard you push (1 mark).
Mark Lesson Complete
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