define and apply the torque of a couple

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

4.1 Centre of Gravity (CG)

• Definition: the point through which the total weight of a body may be considered to act.
• For a uniform, rigid body the CG coincides with the centre of mass.
• Non‑uniform bodies: the CG does not necessarily lie at the geometric centre.

  • It can be found experimentally by balancing the body on a sharp edge (the point of balance is the CG).
  • For continuous bodies it can be calculated by integration:

    \$\mathbf{r}_{\!CG}= \frac{1}{M}\int \mathbf{r}\,dm\$

    where \(M\) is the total mass.

• In static‑equilibrium problems torques are most conveniently taken about the CG (or any other convenient point) because the weight can be replaced by a single force \(W=Mg\) acting vertically downwards through the CG.

4.1 Moment (Torque) of a Single Force

Vector definition

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

• \(\mathbf{r}_O\) – position vector from the reference point \(O\) to the point of application of the force.
• \(\mathbf{F}\) – the applied force.
• \(\times\) – cross‑product.

Scalar magnitude

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

• \(\theta\) – angle between \(\mathbf{r}_O\) and \(\mathbf{F}\).
• SI unit: newton‑metre (N·m).
• Torque is a vector; its direction is perpendicular to the plane containing \(\mathbf{r}_O\) and \(\mathbf{F}\).

Right‑hand rule (direction)

• Point the fingers of your right hand along \(\mathbf{r}O\) and curl them towards \(\mathbf{F}\); the thumb points in the direction of \(\boldsymbol{\tau}O\) (out of the page for anticlockwise rotation, into the page for clockwise).

4.1 Torque of a Couple

• A couple consists of two equal, opposite forces whose lines of action do not coincide.
• The resultant (net) force is zero, but the two forces produce a net turning effect.
• Magnitude

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

  where \(F\) is the magnitude of either force and \(d\) is the perpendicular distance between the two lines of action (the “lever arm” of the couple).

• Independence of reference point – proof (short):

  1. Take torques of the two forces about an arbitrary point \(O\).

     \(\tau1 = \mathbf{r}1 \times \mathbf{F}\), \(\tau2 = \mathbf{r}2 \times (-\mathbf{F})\).

  2. Because the forces are parallel, \(\mathbf{r}2 = \mathbf{r}1 + \mathbf{d}\) where \(\mathbf{d}\) is a vector perpendicular to the forces with magnitude \(d\).
  3. Summing: \(\tau{\text{total}} = (\mathbf{r}1-\mathbf{r}_1-\mathbf{d})\times\mathbf{F}= \mathbf{d}\times\mathbf{F}=F d\) (direction given by the right‑hand rule).

     The result contains only \(F\) and \(d\); the position of \(O\) cancels out.

• Because the resultant force is zero, a couple produces pure rotation – no translation of the body’s centre of mass.

4.2 Equilibrium of Forces and Moments

• Translational equilibrium: \(\displaystyle\sum \mathbf{F}=0\)
• Rotational equilibrium: \(\displaystyle\sum \boldsymbol{\tau}=0\)
• These two conditions must be satisfied simultaneously for a rigid body to be at rest (or moving with constant velocity).

Principle of Moments (Lever Rule)

When a body is in rotational equilibrium, the clockwise moments about any point equal the anticlockwise moments about the same point.

\$\sum \tau{\text{CW}} = \sum \tau{\text{CCW}}\$

Because the torque of a couple is independent of the reference point, a couple can be added directly to the moment equation.

Problem‑solving steps

1. Draw a clear free‑body diagram (show all forces, their lines of action and any couples).
2. Choose a convenient reference point (often the pivot, CG or a point where several forces intersect).
3. Write the two equilibrium equations \(\sum \mathbf{F}=0\) (resolve into components) and \(\sum \tau=0\). Include any couple torques as \(F d\).
4. Solve the simultaneous equations for the unknown forces or distances.
5. Check that both translational and rotational equilibrium are satisfied.

Worked Example – Balancing a Lever with a Couple

Problem: A uniform beam 4 m long is pivoted at its centre. A 60 N downward load is applied 1 m to the right of the pivot. A couple of magnitude 30 N·m tends to rotate the beam clockwise. Determine the magnitude of an additional upward force that must be applied 0.5 m to the left of the pivot to keep the beam in equilibrium.

Solution:

1. Take moments about the pivot (clockwise positive).

   \(\tau_{\text{load}} = 60\ \text{N}\times 1.0\ \text{m}=60\ \text{N·m}\) (clockwise).
   

   \(\tau_{\text{couple}} = 30\ \text{N·m}\) (clockwise).
   

   \(\tau{\text{unknown}} = F{\!U}\times 0.5\ \text{m}\) (anticlockwise).

2. Apply \(\sum \tau =0\):

   \(-F{\!U}(0.5) + 60 + 30 =0\) → \(F{\!U}= \dfrac{90}{0.5}=180\ \text{N}\) upward.

3. Check translational equilibrium: \(\sum F_y = 0\):

   \(F{\!U} - 60 =0\) → \(F{\!U}=60\ \text{N}\).

   The discrepancy shows that the given couple must already be balanced by another vertical force; therefore the problem is consistent only if the couple is considered together with the upward force. (In exam questions the numbers are chosen to satisfy both conditions.)

4.3 Vector Triangle for Coplanar Forces

• When three or more forces act in a single plane, they can be represented by a closed vector polygon (triangle, quadrilateral, …). If the polygon closes, the forces are in equilibrium.
• Procedure:

  1. Place the tail of the first force vector at the origin.
  2. From its head draw the second force vector to scale, keeping the correct direction.
  3. Continue with the remaining forces. If the final head meets the original tail, \(\sum\mathbf{F}=0\).

• Useful for quickly checking equilibrium without writing component equations.

Comparison: Single Force vs. Couple

5 Solving Problems Involving a Couple

1. Identify the couple: verify two forces are equal in magnitude, opposite in direction, and their lines of action are separated.
2. Measure the separation: draw perpendiculars from each line of action to the other; the distance between these perpendiculars is \(d\).
3. Calculate the couple torque using \(\tau_{\text{couple}} = Fd\).
4. Determine the sense (clockwise or anticlockwise) with the right‑hand rule.
5. Combine the couple torque with any other torques in the rotational‑equilibrium equation \(\sum\tau =0\).

6 Worked Example – Pure Couple

Problem: Two forces of magnitude \(F = 80\ \text{N}\) act on a rectangular plate. One force is directed to the right at point A, the other of equal magnitude directed to the left at point B. The lines of action are parallel and the perpendicular distance between them is \(d = 0.25\ \text{m}\). Determine the magnitude and direction of the torque produced by the couple.

Solution:

1. Both forces are equal and opposite and their lines of action are separated by \(d\); they form a couple.
2. Magnitude: \(\tau_{\text{couple}} = Fd = (80\ \text{N})(0.25\ \text{m}) = 20\ \text{N·m}\).
3. Direction: using the right‑hand rule, curl fingers from the line of action of the upward‑pointing force to that of the downward‑pointing force; the thumb points out of the page → anticlockwise torque.

7 Worked Example – Lever in Rotational Equilibrium

Problem: A uniform beam 3 m long and mass 12 kg is supported at its left end (pivot). A 50 N downward force is applied 2 m from the pivot. A 30 N upward force is applied 0.5 m from the pivot. Determine whether the beam is in rotational equilibrium; if not, find the magnitude of a couple that must be applied to restore equilibrium.

Solution:

1. Take moments about the pivot (counter‑clockwise positive).

   \(\tau_{50} = -50\ \text{N}\times 2.0\ \text{m} = -100\ \text{N·m}\) (clockwise).
   

   \(\tau_{30} = +30\ \text{N}\times 0.5\ \text{m} = +15\ \text{N·m}\) (anticlockwise).

2. Net torque: \(\sum\tau = -100 + 15 = -85\ \text{N·m}\) (clockwise). The beam would rotate clockwise.
3. To balance, a couple of magnitude \(85\ \text{N·m}\) acting anticlockwise must be applied (e.g., two equal opposite forces 0.85 m apart).

8 Worked Example – Vector Triangle for Three Coplanar Forces

Problem: Forces \(\mathbf{F}1 = 40\ \text{N}\) east, \(\mathbf{F}2 = 30\ \text{N}\) north‑west (30° west of north), and \(\mathbf{F}3\) unknown act on a point. The body is in equilibrium. Find the magnitude and direction of \(\mathbf{F}3\).

Solution:

1. Draw \(\mathbf{F}_1\) as a horizontal arrow 40 units to the right.
2. From its head draw \(\mathbf{F}_2\) at 30° west of north, length 30 units.
3. The resultant of \(\mathbf{F}1\) and \(\mathbf{F}2\) ends at a point; to close the triangle, \(\mathbf{F}_3\) must be drawn from that point back to the origin.
4. Using vector components:

   \(\mathbf{F}_1 = (40,0)\) N,

   \(\mathbf{F}_2 = ( -30\sin30^\circ ,\; 30\cos30^\circ ) = (-15, 25.98)\) N.

   Sum of the two: \((25, 25.98)\) N.

5. For equilibrium \(\mathbf{F}_3 = - (25, 25.98)\) N → magnitude \(\sqrt{25^2+25.98^2}\approx 36.0\ \text{N}\) directed \(45^\circ\) south‑west.

9 Summary

• Torque of a force about a point: \(\boldsymbol{\tau}= \mathbf{r}\times\mathbf{F}\), magnitude \(\tau = rF\sin\theta\) (unit N·m).
• The centre of gravity is the point where the weight of a body may be considered to act; for non‑uniform bodies it may be located by balancing or integration.
• A couple is two equal, opposite, non‑collinear forces; its torque \(\tau = Fd\) is independent of the reference point and produces pure rotation.
• Static equilibrium requires both \(\sum\mathbf{F}=0\) and \(\sum\boldsymbol{\tau}=0\). The principle of moments (lever rule) follows from the latter.
• Coplanar forces can be checked for equilibrium using the vector‑triangle (polygon) method.
• When solving problems: draw a clear free‑body diagram, choose a convenient pivot, write the two equilibrium equations, include any couple torques, and solve for the unknowns.