understand that the weight of an object may be taken as acting at a single point known as its centre of gravity

Turning Effects of Forces

Objective

Understand that the weight of an object may be taken as acting at a single point known as its centre of gravity (CG) and use this concept to analyse turning (torque) effects.

Key Definitions

• Torque (Turning Effect): The tendency of a force to rotate an object about an axis. Measured in newton‑metres (N·m).
• Lever Arm (Moment Arm): The perpendicular distance between the line of action of the force and the axis of rotation.
• Centre of Gravity (CG): The point at which the total weight of a body may be considered to act.
• Centre of Mass (CM): The point at which the total mass of a body may be considered to act. For uniform gravity fields, CG and CM coincide.

Torque Formula

The magnitude of the torque \$\tau\$ produced by a force \$F\$ applied at a distance \$r\$ from the axis of rotation is

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

where \$\theta\$ is the angle between the force vector and the lever arm. For forces acting perpendicular to the lever arm, \$\sin\theta = 1\$ and the formula simplifies to \$\tau = rF\$.

Why the Weight Can Be Treated at the Centre of Gravity

1. Weight is a uniform force field (gravity) acting on every element of mass.
2. When the gravitational field is parallel and uniform, the resultant of all the individual weight forces passes through a single point – the centre of gravity.
3. This allows us to replace the distributed weight by a single force \$W = mg\$ acting at the CG, simplifying torque calculations.

Determining the Centre of Gravity

For simple geometric bodies the CG can be found by symmetry:

• Uniform rod: midpoint.
• Uniform rectangular plate: intersection of its diagonals.
• Uniform solid sphere or cylinder: geometric centre.

For irregular bodies, the CG can be located experimentally by suspending the object from a point and drawing a vertical line through the suspension point; repeat from another point – the intersection of the two lines is the CG.

Application Example

Consider a uniform beam of length \$L = 4.0\ \text{m}\$ and mass \$m = 20\ \text{kg}\$ supported at its left end (pivot). A load of \$10\ \text{kg}\$ is placed \$1.5\ \text{m}\$ from the pivot. Determine the torque about the pivot due to the beam’s weight and the load.

The total clockwise torque about the pivot is \$\tau_{\text{total}} = 392.4 + 147.15 = 539.55\ \text{N·m}\$.

Practice Questions

1. A uniform rectangular plate of mass \$5\ \text{kg}\$ measures \$0.8\ \text{m} \times 0.6\ \text{m}\$. It is supported at one corner. Calculate the torque due to its weight about the support point.
2. A ladder of length \$3.0\ \text{m}\$ and mass \$12\ \text{kg}\$ leans against a smooth wall, making an angle of \$60^\circ\$ with the ground. The foot of the ladder is \$1.0\ \text{m}\$ from the wall. Determine the torque about the foot due to the ladder’s weight.
3. Explain how the centre of gravity shifts if a non‑uniform object (e.g., a hammer) is held by its handle versus by its head.

Summary

• The torque produced by a force depends on both its magnitude and the perpendicular distance from the axis of rotation.
• In a uniform gravitational field, the weight of an object can be treated as a single force acting at the centre of gravity.
• Identifying the CG simplifies analysis of turning effects, allowing straightforward calculation of torques.