A4-1. Rigid Body Mechanics: Torque and EquilibriumHL ONLY

1. Torque & Couples

A moment is the turning effect of a force around a particular point or pivot. While moments describe turning effects in static situations, torque describes the change in rotational motion due to a turning force.

Torque ($\tau$):

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

Vector Definition:
Torque is a vector quantity defined mathematically as the cross product of the position vector ($\vec{r}$) and the force vector ($\vec{F}$). The direction of the torque vector is determined by the right-hand rule, perpendicular to the plane containing $\vec{r}$ and $\vec{F}$.

$$\vec{\tau} = \vec{r} \times \vec{F}$$
  • $\tau$: Torque in Newton-metres (N m)
  • $F$: Applied force (N)
  • $r$: Distance from the axis of rotation to the point of force application (m)
  • $\theta$: Angle between the force vector and the radial line from the axis.

Note: Maximum torque occurs when the force is applied completely perpendicular to the object ($\theta = 90^\circ$, so $\sin 90^\circ = 1$). If you push directly towards or away from the pivot ($\theta = 0^\circ$), the torque is zero.

Couples:

A couple is a pair of equal and opposite coplanar forces that act to produce rotation only. They produce a net linear force of zero, meaning the object does not accelerate translationally, but it does experience a net torque.

  • Torque of a Couple: $\tau = F \times d$ (where $d$ is the perpendicular distance between the lines of action of the two forces).

Example 1: Torque on a Wrench

Problem: A mechanic applies a 150 N force at an angle of 30° to a wrench. The force is applied 0.24 m from the center of the bolt. Calculate the torque.

r = 0.24 m F = 150 N 30°

Solution: Use the torque equation factoring in the angle:
$$\tau = Fr \sin\theta = (150)(0.24) \sin(30^\circ) = \mathbf{18.0\text{ N m}}$$

2. Rotational Equilibrium

A system is in complete rotational equilibrium when it has no linear acceleration and no angular acceleration. This requires two conditions to be met simultaneously:

  1. Translational Equilibrium: The net resultant force is zero ($\Sigma F = 0$).
  2. Rotational Equilibrium: The net resultant torque is zero ($\Sigma \tau = 0$). The sum of clockwise torques exactly equals the sum of anti-clockwise torques.

Example 2: Unbalanced Torque on a Wheel

Problem: Three forces act on a wheel. A 10 N force acts clockwise at $r = 0.25\text{ m}$. A 9 N force acts clockwise at $r = 0.25\text{ m}$. A 12 N force acts anti-clockwise at an angle of 30° at $r = 0.1\text{ m}$. Determine the net resultant torque and the direction of angular acceleration.

Solution:

  • 1. Sum the Clockwise Torques:
    $\tau_{cw} = (10 \times 0.25 \times \sin 90^\circ) + (9 \times 0.25 \times \sin 90^\circ) = 2.5 + 2.25 = 4.75\text{ N m}$
  • 2. Sum the Anti-Clockwise Torques:
    $\tau_{acw} = 12 \times 0.1 \times \sin 30^\circ = 12 \times 0.1 \times 0.5 = 0.6\text{ N m}$
  • 3. Determine Net Torque:
    $\Sigma \tau = \tau_{cw} - \tau_{acw} = 4.75 - 0.6 = \mathbf{4.15\text{ N m}}$
    Because the clockwise torque is larger, the angular acceleration will be in the clockwise direction.