where "$\times$" denotes the vector cross‑product. The magnitude is
$$\tau = r\,F\,\sin\theta$$
with $\theta$ the angle between $\mathbf{r}$ and $\mathbf{F}$. The direction of $\boldsymbol{\tau}$ follows the right‑hand rule.
Key Points
Moment is a vector quantity; its direction indicates the axis of rotation.
SI unit: newton‑metre (N·m). In practice, the sign (+/–) is used to denote clockwise or anticlockwise rotation.
Only the component of the force perpendicular to the lever arm contributes to the moment.
A couple consists of two equal and opposite forces whose lines of action do not coincide; it produces a pure turning effect with no resultant force.
Sign Convention
For problems in a vertical plane:
Counter‑clockwise moments are taken as positive (+).
Clockwise moments are taken as negative (–).
Resultant Moment
The net turning effect about an axis is the algebraic sum of all individual moments:
$$\sum \tau = \tau_1 + \tau_2 + \dots + \tau_n$$
For rotational equilibrium (no angular acceleration), the condition is
$$\sum \tau = 0$$
Table of Symbols
Symbol
Quantity
Unit
Definition
$\tau$
Moment (torque)
N·m
Turning effect of a force about an axis
$\mathbf{F}$
Force
N
External influence causing linear acceleration
$\mathbf{r}$
Position vector (lever arm)
m
Distance from axis to point of application of $\mathbf{F}$
$\theta$
Angle between $\mathbf{r}$ and $\mathbf{F}$
degrees or radians
Determines the perpendicular component of the force
Worked Example
Problem: A uniform beam 4.0 m long is hinged at its left end (point A) and supported by a rope at its right end (point B). The beam carries a load of 800 N at its centre. The rope makes an angle of $30^{\circ}$ above the horizontal. Determine the tension in the rope.
Choose the hinge A as the axis of rotation. This eliminates the unknown hinge reaction forces from the moment equation.
Calculate the perpendicular distances:
Weight $W = 800\,$N acts at $2.0\,$m from A.
The tension $T$ acts at $4.0\,$m from A, with a vertical component $T\sin30^{\circ}$ producing a clockwise moment.
Write the moment equilibrium about A (counter‑clockwise positive):
$$\sum \tau_A = 0 = W(2.0) - T\sin30^{\circ}(4.0)$$
Solve for $T$:
$$800 \times 2.0 = T \times \frac{1}{2} \times 4.0$$
$$1600 = 2.0\,T$$
$$T = 800\ \text{N}$$
The tension in the rope is $800\,$N.
Common Mistakes
Using the full magnitude of the force instead of its component perpendicular to the lever arm.
Neglecting the sign convention, leading to incorrect addition/subtraction of moments.
Forgetting to include all forces that produce a moment about the chosen axis.
Suggested Diagram
Suggested diagram: Beam hinged at A, rope at B making $30^{\circ}$ with the horizontal, and a downward load at the centre. Indicate lever arms and direction of moments.
Practice Questions
A force of $120\,$N is applied at the end of a $0.5\,$m long wrench, making an angle of $45^{\circ}$ with the wrench. Calculate the moment about the bolt.
A uniform rectangular plate $0.8\,$m by $0.6\,$m rests on a frictionless pivot at its left edge. A force of $50\,$N is applied at the top right corner vertically downwards. Determine the angular acceleration if the plate’s moment of inertia about the pivot is $0.12\,$kg·m².
Two forces of $30\,$N and $40\,$N act on a door at distances of $0.6\,$m and $0.9\,$m from the hinges, respectively, both perpendicular to the door. What is the net moment about the hinges? Is the door in rotational equilibrium?
Summary
The moment of a force quantifies its turning effect about an axis and is given by the cross‑product $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$. Correctly identifying lever arms, perpendicular components, and applying the sign convention are essential for solving A‑Level physics problems involving rotational equilibrium and dynamics.