understand that a couple is a pair of forces that acts to produce rotation only

Turning Effects of Forces (Cambridge IGCSE/A‑Level 9702)

Learning Objectives

• Define the centre of gravity (CG) and justify treating the weight of a rigid body as acting at the CG.
• Derive and use the torque (moment) formula for a single force about any point or any axis.
• State and apply the principle of moments (Σ τ = 0) together with Σ F = 0 for static equilibrium.
• Explain what a couple is, why its resultant force is zero, and prove that its torque is independent of the reference point.
• Calculate the resultant torque of a system of several forces by vector addition.

Key Concepts

• Centre of gravity (CG) – the point at which the total weight of a body may be considered to act.
• Line of action – the straight line along which a force acts.
• Torque (moment) of a force – the turning effect of a force about a chosen point or axis; vector quantity  \(\boldsymbol{\tau}= \mathbf r \times \mathbf F\).
• Resultant force – the vector sum of all forces acting on a body (Σ F).
• Resultant torque – the vector sum of all torques about a chosen point or axis (Σ τ).
• Couple – a pair of equal, opposite, parallel forces whose lines of action are not collinear.
• Static equilibrium – a state in which a rigid body has no translational or rotational acceleration, i.e. Σ F = 0 and Σ τ = 0.

Sign‑Convention Reminder

1. Centre of Gravity

The centre of gravity (CG) of a rigid body is the point at which the total weight \(W=mg\) can be assumed to act. For a uniform (homogeneous) body the CG coincides with the geometric centre; for irregular bodies it is found by balancing moments (the principle of moments).

2. Torque (Moment) of a Single Force

For a force \(\mathbf F\) acting at point A, the torque about a reference point O (or about any axis through O) is defined by the cross‑product

\[ \boldsymbol{\tau}= \mathbf r_{OA}\times\mathbf F, \] where \(\mathbf r_{OA}\) is the position vector from O to the point of application A.

The magnitude is

\[ \tau = r\,F\sin\theta, \] with \(\theta\) the angle between \(\mathbf r\) and \(\mathbf F\). If the axis of rotation is specified (e.g. the z‑axis), only the component of \(\boldsymbol{\tau}\) along that axis is relevant.

Moment About Any Axis

When the axis is not perpendicular to the plane of \(\mathbf r\) and \(\mathbf F\), the torque about the axis \(\mathbf{\hat n}\) is

\[ \tau_{(\mathbf{\hat n})}= (\mathbf r\times\mathbf F)\cdot\mathbf{\hat n}. \]

Thus the same formula works for a point (any axis through the point) or for a fixed axis not passing through the point.

3. Resultant Force of a System of Forces

If several forces \(\mathbf F_1,\mathbf F_2,\dots,\mathbf F_n\) act on a body, the resultant (net) force is the vector sum

\[ \mathbf R = \sum_{i=1}^{n}\mathbf F_i. \]

Only when \(\mathbf R= \mathbf 0\) does the body experience no translational acceleration. Translational equilibrium (Σ F = 0) must be satisfied before considering rotational equilibrium.

4. Couples – Pure Rotation

A couple consists of two forces \(\mathbf F_1\) and \(\mathbf F_2\) such that

1. \(\mathbf F_1 = -\mathbf F_2\) (equal magnitude, opposite direction),
2. The forces are parallel, and
3. Their lines of action are separated by a perpendicular distance \(d\).

Because the forces are equal and opposite, the resultant force is zero:

\[ \mathbf R = \mathbf F_1+\mathbf F_2 = \mathbf 0, \] so the body does not translate. However the pair produces a non‑zero torque (a pure turning effect)

\[ \boxed{\;\tau_{\text{couple}} = F\,d\;} \] directed perpendicular to the plane containing the forces (right‑hand rule).

Proof of Reference‑Point Independence

Let the forces act at points A and B with position vectors \(\mathbf r_A\) and \(\mathbf r_B\) measured from an arbitrary origin O. The torques are

\[ \boldsymbol{\tau}_A = \mathbf r_A \times \mathbf F,\qquad \boldsymbol{\tau}_B = \mathbf r_B \times (-\mathbf F). \]

The resultant torque of the couple about O is

\[ \boldsymbol{\tau}_{\text{couple}} = \boldsymbol{\tau}_A+\boldsymbol{\tau}_B = (\mathbf r_A-\mathbf r_B)\times\mathbf F. \]

The vector \(\mathbf r_A-\mathbf r_B\) is exactly the displacement from B to A, whose magnitude is the perpendicular separation \(d\) and whose direction is normal to the plane of the forces. Hence

\[ \boldsymbol{\tau}_{\text{couple}} = d\,F\,\mathbf{\hat n}, \] which contains no reference‑point vector; therefore the torque of a couple is the same about **any** origin O (or any axis parallel to \(\mathbf{\hat n}\)).

Couple vs. General Force Pair

• Force couple: equal, opposite, parallel forces; resultant force = 0; produces a *pure* rotation.
• General force pair: any two forces (not necessarily equal or parallel). The pair may produce both a net force and a net torque; it is **not** a couple unless the two conditions above are satisfied.

5. Resultant Torque of a System of Forces

For forces \(\mathbf F_i\) acting at points with position vectors \(\mathbf r_i\) (relative to a chosen origin O), the total torque is

\[ \boldsymbol{\tau}_{\text{result}} = \sum_{i}\mathbf r_i \times \mathbf F_i. \]

If all torques are parallel to the same axis (e.g. out of the page), they can be added algebraically using the sign convention.

6. Equilibrium of Forces (Principle of Moments)

A rigid body is in static equilibrium when

\[ \sum \mathbf F = \mathbf 0 \qquad\text{and}\qquad \sum \boldsymbol{\tau}= \mathbf 0. \]

Both conditions must be satisfied simultaneously; neglecting either can lead to an incorrect solution.

Worked Example – Beam on a Support (Using Both Σ F = 0 and Σ τ = 0)

A uniform beam 3.0 m long and weight 600 N rests on a hinge at its left end (point A). A tension force of 400 N acts upward at a point 2.0 m from A. Determine the reaction at the hinge.

1. Free‑body diagram – forces: weight \(W\) acting at the centre (1.5 m from A), tension \(T=400\;\text{N}\) upward at 2.0 m, hinge reaction \(\mathbf R=(R_x,R_y)\) at A.
2. Translational equilibrium (Σ F = 0) \[ \begin{cases} \Sigma F_x: & R_x = 0,\\[4pt] \Sigma F_y: & R_y + T - W = 0 \;\Longrightarrow\; R_y = W - T = 600-400 = 200\;\text{N (upward)}. \end{cases} \]
3. Rotational equilibrium (Σ τ = 0) about point A (so \(\mathbf R\) produces no moment) \[ \tau_A = T(2.0) - W(1.5) = 0 \;\Longrightarrow\; 400(2.0) = 600(1.5). \] Both sides equal 800 N·m, confirming the torque balance. Hence the beam is in equilibrium.
4. Result – hinge reaction \(\mathbf R = (0,\;200\;\text{N})\). The example shows why both Σ F = 0 and Σ τ = 0 are required.

7. Comparison – Single Force vs. Couple

Aspect	Single Force	Force Couple
Resultant force	Non‑zero → translation (and possibly rotation)	Zero → no translation
Resultant torque	Depends on the chosen point or axis	Same magnitude and direction about any point/parallel axis
Effect on a rigid body	Translation + possible rotation	Pure rotation
Units	Force: N; Torque: N·m	Torque only: N·m

8. Practice Questions

1. Two forces of \(50\;\text{N}\) act on a rectangular plate, one upward on the left edge and one downward on the right edge, 0.40 m apart. Determine the magnitude and sense (clockwise/anticlockwise) of the torque produced by the couple.
2. A door 0.80 m wide is opened by applying a horizontal force of \(30\;\text{N}\) at the outer edge, perpendicular to the plane of the door.
   1. What torque does this produce about the hinges?
   2. Is this situation a couple? Explain.
3. Show explicitly that the torque of the wrench couple in the worked example is the same about any point by calculating the torque about a point 0.20 m to the right of the centre.
4. A uniform beam of length 3.0 m and weight 600 N rests on a support at its left end and is held in equilibrium by a tension force of 400 N acting upward at a point 2.0 m from the left end. Determine the reaction force at the support and verify that both Σ F = 0 and Σ τ = 0 are satisfied.
5. Three forces act on a flat plate: \(F_1 = 30\;\text{N}\) upward at point A, \(F_2 = 30\;\text{N}\) downward at point B (1.5 m to the right of A), and a horizontal force \(F_3 = 20\;\text{N}\) to the right at point C (0.8 m above A). Find the resultant torque about point A and state whether the plate is in rotational equilibrium.

9. Summary

• The centre of gravity is the point where the weight of a body may be considered to act.
• Torque of a single force is \(\boldsymbol{\tau}= \mathbf r\times\mathbf F\); its magnitude is \(\tau = rF\sin\theta\) and it depends on the chosen point or axis.
• The resultant force of a system is the vector sum Σ F; translational equilibrium requires Σ F = 0.
• A couple consists of two equal, opposite, parallel forces separated by a distance \(d\); its resultant force is zero but its torque \(\tau = Fd\) is the same about any point (proved using the cross‑product).
• The resultant torque of several forces is Σ τ; rotational equilibrium requires Σ τ = 0 about any axis.
• Both Σ F = 0 and Σ τ = 0 must be satisfied simultaneously to solve static‑equilibrium problems.