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 – Centre of Gravity

Objective

Students will be able to:

• Explain why the weight of a body can be replaced by a single force acting at its centre of gravity (CG).
• Define the moment (torque) of a force and state the principle of moments.
• Calculate moments for forces acting at any angle, including the special case of a couple.
• Determine the CG of regular and irregular objects (theory and experiment).

Key Definitions

Principle of Moments (Equilibrium Condition)

For a body to be in rotational equilibrium about a given axis, the algebraic sum of all moments about that axis must be zero:

In practice we assign a sign (e.g., clockwise = –, anticlockwise = +) and add the moments algebraically.

Moment (Torque) Formulae

• Scalar form (perpendicular force): τ = r F
• General scalar form: τ = r F sin θ
• Vector form: τ = r × F (direction given by the right‑hand rule).

Sign Convention

• Choose a convention and use it consistently throughout a problem.
• Typical school convention: clockwise moments = –, anticlockwise moments = +.
• When several moments act about the same axis, add them algebraically to obtain the net moment.

Why the Weight Can Be Treated at the Centre of Gravity

1. Gravity is a uniform field; every infinitesimal mass element dm experiences the same acceleration g.
2. The weight of each element is dm·g and all these forces are parallel.
3. The vector sum of all the elemental weights is a single resultant W = mg acting through the CG.
4. Replacing the distributed weight by this single force greatly simplifies moment calculations without loss of accuracy.

Determining the Centre of Gravity

By symmetry (regular, homogeneous bodies)

• Uniform rod – midpoint.
• Uniform rectangular plate – intersection of the diagonals.
• Uniform solid sphere, cylinder, cube – geometric centre.

Experimentally (irregular bodies – suspension method)

1. Suspend the object from any convenient point so it can hang freely without touching anything else.
2. Draw a vertical line (using a plumb line) through the suspension point; this line passes through the CG.
3. Repeat the procedure from a different suspension point.
4. The intersection of the two vertical lines marks the CG of the object.

Couples

• Moment of a couple: τ_couple = F d, where F is the magnitude of either force and d is the perpendicular distance between their lines of action.
• The moment of a couple is independent of the reference axis – it has the same value about any point.
• Direction of the moment follows the right‑hand rule (or the chosen sign convention).

Worked Example 1 – Simple Moment Calculation

A uniform beam of length L = 4.0 m and mass m = 20 kg is pivoted at its left end. A 10 kg load is attached 1.5 m from the pivot. Determine the net moment about the pivot.

Net moment (using clockwise = –):

The negative sign indicates a clockwise rotation.

Worked Example 2 – Using the Principle of Moments (Equilibrium)

A uniform ladder 3.0 m long and mass 12 kg leans against a smooth vertical wall. The foot of the ladder is 1.0 m from the wall, making an angle of 60° with the ground. The wall exerts a normal reaction R at the top of the ladder. Determine R using the principle of moments about the foot.

• Weight of ladder: W = mg = 12 × 9.81 = 117.7 N, acting at the ladder’s centre (1.5 m from the foot).
• Perpendicular distance of W from the foot‑ground axis = (1.5 m) sin 60° = 1.30 m.
• Reaction R acts horizontally at the top, 3.0 m from the foot. Its perpendicular distance from the foot‑ground axis = (3.0 m) cos 60° = 1.5 m.
• Taking anticlockwise moments as positive:
  

  Σ τ = 0 ⇒ R (1.5) – W (1.30) = 0

• Hence R = W (1.30) / 1.5 = 117.7 × 1.30 / 1.5 ≈ 102 N (anticlockwise).

Practice Questions

1. A uniform rectangular plate of mass 5 kg measures 0.80 m × 0.60 m. It is supported at one corner. Calculate the moment of its weight about the support point.
2. A ladder 3.0 m long and mass 12 kg leans against a smooth wall, making an angle of 60° with the ground. The foot of the ladder is 1.0 m from the wall. Determine the moment about the foot due to the ladder’s weight.
3. Explain how the centre of gravity of a non‑uniform object (e.g., a hammer) shifts when it is held by the handle versus by the head.
4. Two 15 N forces act on a ruler 0.40 m apart, one upward on the left end and one downward on the right end. Find the moment of the couple and state its sense.
5. Using the suspension method, outline a step‑by‑step procedure to locate the CG of an irregularly shaped wooden block.

Summary

• Moment (torque) quantifies the turning effect of a force: τ = r F sin θ (scalar) or τ = r × F (vector).
• In a uniform gravitational field the weight of a body can be replaced by a single force W = mg acting at the centre of gravity.
• The principle of moments (Σ τ = 0) allows us to solve equilibrium problems and to find unknown forces or distances.
• A couple produces a pure rotation; its moment is τ = F d and is independent of the axis chosen.
• Determine the CG by symmetry for regular bodies or experimentally (suspension method) for irregular ones.
• Apply a consistent sign convention when adding moments.