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 (Moments)

Learning Objective

Apply the principle of moments to situations with one force on each side of a pivot (including the balancing of a beam), and demonstrate the required experimental skills (AO3) as set out in the Cambridge IGCSE Physics 0625 syllabus.

Key Concepts

• Force (F) – a push or pull acting at a point (unit: newton, N).
• Pivot (fulcrum) – the fixed point about which a body rotates.
• Lever arm (d) – the perpendicular distance from the line of action of the force to the pivot (unit: metre, m). Important: the distance must be measured at right‑angles to the force direction, not simply along the beam.
• Moment (torque, M) – the turning effect of a force: M = F × d (unit: newton‑metre, N·m).
• Principle of Moments – a rigid body is in rotational equilibrium when the sum of clockwise moments equals the sum of anticlockwise moments about any chosen pivot.
• Rotational equilibrium – the body either remains at rest or rotates at constant angular speed (∑M = 0).

Definition of a Moment

The moment M of a force F about a pivot is

where d is measured perpendicular to the line of action of the force.

Direction of Moments

Principle of Moments

For rotational equilibrium about a chosen pivot:

Balancing a Beam with One Force on Each Side

When a uniform beam is supported at a pivot, a force F₁ at distance d₁ on one side and a force F₂ at distance d₂ on the opposite side will balance if

Simple Seesaw Example (sanity‑check)

A 3.0 m seesaw is pivoted at its centre. A child weighing 250 N sits 0.8 m from the pivot on the left. Where must a child weighing 150 N sit on the right to keep the seesaw level?

1. Write the equilibrium equation: 250 N × 0.8 m = 150 N × d₂.
2. Solve for d₂: d₂ = (250 × 0.8) / 150 = 1.33 m.
3. Therefore the 150 N child must sit 1.33 m from the pivot on the right side.

Worked Example (non‑trivial)

Problem: A uniform 2.4 m beam is pivoted 0.9 m from its left end. A 35 N weight is hung at the left end. At what distance from the right end must a 25 N weight be placed so that the beam remains in equilibrium?

1. Identify pivot and distances

   • Left‑hand moment: d₁ = 0.9 m.
   • Let x be the distance from the right end to the 25 N weight.

     Right‑hand distance from pivot: d₂ = (2.4 − 0.9) − x = 1.5 m − x.

2. Apply the principle of moments (clockwise = anticlockwise)
   +35 N × 0.9 m  −  25 N × (1.5 m − x) = 0

3. Solve for x

   31.5 = 25(1.5 − x) → 31.5 = 37.5 − 25x → 25x = 6.0 → x = 0.24 m

   The 25 N weight must be placed 0.24 m from the right end (i.e. 1.26 m from the pivot).

4. Check – Both moments are 31.5 N·m, confirming equilibrium.

Resultant of Two Perpendicular Forces (Supplementary)

When two forces act at right angles, they can be combined into a single resultant using the Pythagorean theorem:

The direction of R is given by tan θ = F₂ / F₁. In a moment problem the resultant force can be used in place of the two original forces, provided its line of action is correctly identified.

Centre of Gravity & Stability (Brief Note – 1.5.3 Link)

• For a uniform beam the centre of gravity (CG) lies at its geometric centre. The weight of the beam therefore produces no moment about a pivot placed at the CG.
• If the pivot is not at the CG, include the beam’s own weight: M_beam = (mass × g) × distance from pivot to CG.
• Stability requires that the line of action of the resultant force passes through the base of support; otherwise the beam will tip.

Mini‑Investigation (AO3 – Planning, Recording, Evaluating)

Objective: Verify experimentally that F₁ × d₁ = F₂ × d₂ for a balanced beam.

1. Planning

   • Equipment – metre‑rule (or wooden beam), knife‑edge pivot, calibrated masses (0.5 kg, 1 kg, 2 kg), string, ruler, safety goggles.
   • Risk Assessment

     • Knife‑edge is sharp – handle with care.
     • Hanging masses may fall – secure strings and keep hands clear.
     • Ensure the work surface is stable to prevent the beam from sliding.

2. Recording

   • Sketch a labelled diagram showing the beam, pivot, forces and distances.
   • Use the data table below to record each trial.

3. Evaluating

   • Identify sources of error (friction, measurement inaccuracy, beam’s own weight).
   • Suggest improvements (low‑friction pivot, vernier calipers for distance, include beam weight in calculations).

Data Table (example)

Common Mistakes & How to Avoid Them

• Using the total length of the beam instead of the perpendicular distance from the pivot.
• Neglecting a consistent sign convention – always write the sign before each moment term.
• Forgetting the beam’s own weight when the pivot is not at its centre of gravity.
• Mixing units (e.g., cm with m); convert all distances to metres before calculation.

Summary Table

Practice Questions

1. A uniform 1.5 m beam is supported at its centre. A 40 N force is applied 0.3 m to the left of the pivot. Where should a 25 N force be applied on the right side to keep the beam in equilibrium?
2. A seesaw 4.0 m long pivots at its centre. A child weighing 300 N sits 0.9 m from the pivot. How far from the pivot must a second child weighing 200 N sit on the opposite side to balance the seesaw?
3. A metre‑rule (mass 0.5 kg, uniform) is balanced on a knife‑edge at the 30 cm mark. A 2.0 kg mass is hung at the 0 cm end. Calculate the distance from the knife‑edge at which a 1.0 kg mass must be placed on the right side to achieve equilibrium. (Take g = 9.8 m s⁻².)
4. In a laboratory set‑up, a 3.0 kg block is attached to a string that passes over a frictionless pulley and pulls down on the left side of a 1.2 m beam. The pivot is 0.4 m from the left end. If the beam is in equilibrium, what upward force (in N) must be applied at the right end 0.8 m from the pivot?