Apply the principle of moments to situations with one force each side of the pivot, including balancing of a beam

1.5.2 Turning Effect of Forces

What is a Turning Effect?

The turning effect (or moment) of a force is how much that force tends to rotate an object about a pivot point. It depends on two things:

  • the magnitude of the force, \$F\$

  • the perpendicular distance from the pivot to the line of action of the force, \$d\$

Mathematically, the moment is \$M = F \times d\$. The larger the moment, the stronger the tendency to rotate.

Principle of Moments (Leverage)

When a rigid body is in rotational equilibrium (i.e. it does not rotate), the sum of all clockwise moments equals the sum of all counter‑clockwise moments:

\$\sum M{\text{clockwise}} = \sum M{\text{counter‑clockwise}}\$

This is the same idea as balancing a seesaw: the heavier side must be farther from the pivot to stay level.

One Force on Each Side of a Pivot

In many IGCSE problems you’ll see a simple beam with a single force on each side of a pivot. The beam is balanced when:

\$F1 \times d1 = F2 \times d2\$

where \$F1\$ and \$F2\$ are the forces, and \$d1\$ and \$d2\$ are their distances from the pivot.

Analogy: The Seesaw

Imagine a playground seesaw. If you sit 1 m from the centre and your friend sits 2 m away, you need to be twice as light as your friend to stay level. That’s exactly the principle of moments in action!

Example 1: Balancing a Beam

Suppose a beam has a 5 kg weight hanging 0.4 m to the left of the pivot and a 3 kg weight hanging 0.6 m to the right. Does the beam balance?

  1. Convert masses to forces: \$F = mg\$. Take \$g = 9.8\,\text{m/s}^2\$.

  2. Compute left moment: \$M_{\text{L}} = 5\,\text{kg} \times 9.8\,\text{m/s}^2 \times 0.4\,\text{m} = 19.6\,\text{N·m}\$.

  3. Compute right moment: \$M_{\text{R}} = 3\,\text{kg} \times 9.8\,\text{m/s}^2 \times 0.6\,\text{m} = 17.64\,\text{N·m}\$.

  4. Compare: \$19.6\,\text{N·m} \neq 17.64\,\text{N·m}\$, so the beam tips to the left.

To balance, you could move the 3 kg weight further right or add a small weight on the left.

Example 2: Finding an Unknown Force

A beam is balanced with a 4 kg weight 0.5 m to the left of the pivot. An unknown weight \$x\$ is 0.3 m to the right. What is \$x\$?

SideMass (kg)Distance (m)Moment (N·m)
Left40.5\$4 \times 9.8 \times 0.5 = 19.6\$
Right\$x\$0.3\$x \times 9.8 \times 0.3\$
Set moments equal\$19.6 = 2.94x\$
Solve for \$x\$\$x = \dfrac{19.6}{2.94} \approx 6.67\$ kg

Practice Problems

  1. A 2 kg weight hangs 0.8 m to the left of a pivot. A 3 kg weight hangs 0.4 m to the right. Is the beam balanced? If not, which side tips?

  2. Find the distance from the pivot that a 5 kg weight must be placed on the right side to balance a 10 kg weight 0.3 m to the left.

  3. On a beam, a 4 kg weight is 0.6 m to the left. A 2 kg weight is 0.4 m to the right. What additional mass must be added on the right side, 0.2 m from the pivot, to achieve equilibrium?

Key Takeaways

  • Moment = Force × Distance.

  • For equilibrium: sum of clockwise moments = sum of counter‑clockwise moments.

  • Changing either the force or the distance can balance a beam.

  • Use the seesaw analogy to visualise how distance compensates for weight.

Keep practising, and soon you’ll be able to solve any moment problem with confidence! 🚀