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

What is the Centre of Gravity?

The centre of gravity (CG) is the point at which the weight of an object can be considered to act.

Mathematically, if an object has a total weight \$W = mg\$, the CG is the point where the vector sum of all weight forces is equivalent to a single force \$W\$ acting at that point.

This simplification lets us treat complex shapes as if all their mass were concentrated at one spot.

Why Can We Treat Weight as Acting at a Single Point?

Because weight is a uniform field acting downwards, the torque produced by the weight about any pivot is the same whether the weight is distributed or concentrated at the CG.

The torque is given by \$\tau = \mathbf{r} \times \mathbf{F}\$

where \$\mathbf{r}\$ is the position vector from the pivot to the CG and \$\mathbf{F}\$ is the weight force.

If we replace the distributed weight by a single force \$W\$ at the CG, the same \$\tau\$ results.

Analogy: The Balance Scale ⚖️

Imagine a balance scale with a beam and two pans.

If you put a heavy book on one pan, the scale tips because the book’s weight acts at the point where it touches the pan.

Even if the book were a long, uneven shape, the scale behaves the same as long as we consider the weight acting at the point where the book’s centre of gravity touches the pan.

This is exactly what we do with any object: we look at the point where its CG touches the support.

Example: A Book on a Table 📚

A 0.5 kg book sits on a table.

Its weight is \$W = 0.5\,\text{kg} \times 9.8\,\text{m/s}^2 = 4.9\,\text{N}\$.

The CG of the book is at its centre, so the weight acts at the centre.

If the book is 0.3 m long, the torque about the table edge (pivot) is

\$\tau = r \times W = 0.15\,\text{m} \times 4.9\,\text{N} = 0.735\,\text{N·m}.\$

If we replace the book by a 4.9 N force acting at the centre, we get the same torque.

Exam Tip Box 📝

Tip: When a question asks for the turning effect of a weight, always locate the CG first.

Use the formula \$\tau = r \times F\$ and remember that \$r\$ is the perpendicular distance from the pivot to the line of action of \$F\$.

If the weight is distributed, find the CG and treat the weight as a single force at that point.

Practice Questions

1. A 2 kg block of uniform density is 0.4 m long. It rests on a horizontal surface. What is the torque about the left edge if a 10 N force is applied at the right edge?
2. Explain why a long, unevenly shaped object can be treated as if its weight acts at its centre of gravity when calculating turning effects.
3. Draw a diagram (described in words) showing the CG of a rectangular plate and the torque produced by its weight about a pivot at one corner.

Answers

1. Torque \$\tau = r \times F = 0.4\,\text{m} \times 10\,\text{N} = 4\,\text{N·m}\$ (counter‑clockwise).
2. Because the weight is a uniform field, the net torque depends only on the CG. Replacing the distributed weight by a single force at the CG yields the same torque about any pivot.
3. Describe a rectangle with sides \$a\$ and \$b\$. The CG is at \$(a/2,\,b/2)\$. If the pivot is at one corner, the distance \$r = \sqrt{(a/2)^2 + (b/2)^2}\$. The torque is \$r \times mg\$ with direction given by the right‑hand rule.

Summary Table of Key Concepts