define and apply the moment of a force

Turning Effects of Forces (Cambridge IGCSE/A‑Level Physics 9702 – Topic 4.1)

Learning Objective

Define the moment (torque) of a force, understand its vector nature, and apply the principle of moments to solve quantitative problems involving rotational equilibrium and, where relevant, rotational dynamics.

1. Definition and Vector Nature of a Moment

  • Moment (torque) of a force about a chosen axis is the tendency of that force to produce rotation about that axis.

  • Mathematically it is the vector cross‑product

    \$\boldsymbol{\tau}= \mathbf{r}\times\mathbf{F}\$

    where r is the position vector from the axis to the point of application of the force F.

  • Magnitude:

    \$\tau = r\,F\sin\theta\$

    with θ the angle between r and F. Only the component of the force perpendicular to the lever arm (\(F_{\perp}=F\sin\theta\)) contributes.

  • Direction (right‑hand rule): point the fingers of your right hand along r, curl them toward F; the thumb points in the direction of \(\boldsymbol{\tau}\). In 2‑D diagrams this is shown as an arrow out of the page (counter‑clockwise) or into the page (clockwise).

2. Units and Sign Convention

  • SI unit: newton‑metre (N·m). (Although 1 N·m = 1 J, torque is a vector while energy is a scalar.)

  • Sign convention (choose once and keep it):

    • Counter‑clockwise moments are taken as positive (+).

    • Clockwise moments are taken as negative (–).

3. Principle of Moments (Rotational Equilibrium)

  • The algebraic sum of all moments about any axis is

    \$\displaystyle\sum\tau = \tau{1}+\tau{2}+…+\tau_{n}\$

  • If the body is in rotational equilibrium (no angular acceleration),

    \$\boxed{\displaystyle\sum\tau = 0}\$

  • For two forces acting about the same axis the equilibrium condition reduces to the familiar lever law:

    \$\boxed{r{1}F{1}=r{2}F{2}}\$

    where \(r{1},r{2}\) are the perpendicular distances (lever arms) from the axis to the lines of action of the forces.

4. Couples (Pure Turning Effect)

  • A couple consists of two equal and opposite forces whose lines of action do not coincide.

  • The resultant force of a couple is zero, but the resultant torque is non‑zero:

    \$\tau_{\text{couple}} = 2F\,d\$

    where \(d\) is the perpendicular distance between the two forces.

  • Because a couple produces a torque without a net force, it causes pure rotation – a useful concept for tools such as wrenches.

5. Rotational Dynamics (A‑Level Extension)

  • When angular acceleration \(\alpha\) is present, the net torque about an axis is related to the moment of inertia \(I\) of the body:

    \$\boxed{\displaystyle\sum\tau = I\alpha}\$

  • This relation is not required for IGCSE but appears in the A‑Level syllabus and links torque to rotational motion.

6. Practical Tips for Solving Torque Problems

  • Choose the axis wisely – a hinge, pivot, or any point that makes one or more unknown forces pass through the axis (zero torque).

  • Identify the perpendicular component of each force: use \(F\sin\theta\) or resolve the force into components.

  • Apply the sign convention consistently – write each moment with its sign before summing.

  • Remember: a force whose line of action passes through the chosen axis contributes zero torque.

7. Table of Symbols

SymbolQuantityUnitDefinition
\(\tau\)Moment (torque)N·mTurning effect of a force about an axis
\(\mathbf{F}\)ForceNExternal influence causing linear acceleration
\(\mathbf{r}\)Position vector (lever arm)mDistance from the chosen axis to the point of application of \(\mathbf{F}\)
\(\theta\)Angle between \(\mathbf{r}\) and \(\mathbf{F}\)rad (or °)Determines the perpendicular component of the force
\(I\)Moment of inertiakg·m²Rotational analogue of mass
\(\alpha\)Angular accelerationrad·s⁻²Rate of change of angular velocity

8. Worked Example – Beam with a Rope (Rotational Equilibrium)

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). A downward load of 800 N acts at the centre of the beam. The rope makes an angle of \(30^{\circ}\) above the horizontal. Determine the tension \(T\) in the rope.

  1. Choose the hinge A as the axis. Forces whose lines of action pass through A (the hinge reaction) give zero torque.

  2. Calculate the perpendicular components:

    • Weight \(W = 800\text{ N}\) acts 2.0 m from A. Its line of action is vertical, so the moment is \(+W(2.0)\) (counter‑clockwise).

    • Tension \(T\) acts at 4.0 m from A. Only the vertical component \(T\sin30^{\circ}\) creates a moment, producing a clockwise (negative) torque: \(-T\sin30^{\circ}(4.0)\).

  3. Write the moment equilibrium about A (counter‑clockwise positive):

    \$\sum\tau_{A}=0 \;\Rightarrow\; +W(2.0)\;-\;T\sin30^{\circ}(4.0)=0\$

  4. Insert the numbers and solve:

    \[

    800\times2.0 \;-\; T\left(\frac12\right)(4.0)=0\;\Longrightarrow\;1600 = 2.0\,T\;\Longrightarrow\;T = 800\text{ N}

    \]

Hence the tension in the rope is 800 N.

9. Worked Example – A Simple Couple

Problem: Two opposite forces of 50 N are applied to a wrench 0.30 m apart (perpendicular distance). Find the torque produced by the couple.

  • Because the forces are equal and opposite, the net force is zero, but the torque is

    \[

    \tau_{\text{couple}} = 2F\,d = 2(50\text{ N})(0.30\text{ m}) = 30\text{ N·m}

    \]

    (counter‑clockwise taken as positive).

The wrench will rotate as if a single 30 N·m torque were applied.

10. Common Mistakes & How to Avoid Them

  • Using the full force magnitude instead of the perpendicular component – always multiply by \(\sin\theta\) or work directly with the resolved component.

  • Neglecting the sign convention – decide whether counter‑clockwise is + or – before writing the moment equation and keep it consistent.

  • Forgetting that forces through the axis give zero torque – this is a powerful way to eliminate unknown reactions.

  • Confusing torque with energy – torque is a vector (has direction); energy is a scalar (no direction), even though both use N·m.

  • Omitting a couple’s contribution – remember that a pair of equal opposite forces can produce a net torque even when the resultant force is zero.

11. Practice Questions

  1. A force of 120 N is applied at the end of a 0.50 m wrench, making an angle of \(45^{\circ}\) with the wrench. Calculate the moment about the bolt.

  2. A uniform rectangular plate \(0.80\text{ m}\times0.60\text{ m}\) rests on a frictionless pivot at its left edge. A force of 50 N is applied at the top‑right corner vertically downwards. The plate’s moment of inertia about the pivot is \(0.12\text{ kg·m}^2\). Determine the angular acceleration \(\alpha\).

  3. Two forces of 30 N and 40 N act on a door at distances of 0.60 m and 0.90 m from the hinges, respectively, both perpendicular to the door. What is the net moment about the hinges? Is the door in rotational equilibrium?

  4. A pair of opposite forces each of magnitude 25 N act on a flat bar, the lines of action being 0.40 m apart. Find the torque of the couple and state the direction (choose a sign convention).

12. Summary

The moment of a force quantifies its turning effect about an axis and is given by the vector cross‑product \(\boldsymbol{\tau}= \mathbf{r}\times\mathbf{F}\). Its magnitude depends on the perpendicular component of the force, \(\tau = rF\sin\theta\). By adopting a consistent sign convention, recognising that forces whose lines of action pass through the chosen axis give zero torque, and applying the equilibrium condition \(\sum\tau = 0\) (or the lever law \(r{1}F{1}=r{2}F{2}\)), you can solve all Cambridge IGCSE/A‑Level problems involving rotational equilibrium. For A‑Level, extend the analysis with \(\sum\tau = I\alpha\) to relate torque to angular acceleration.