Explain why the metre rule is first balanced at its centre of gravity before any masses are hung on it.

With the pivot at the centre of gravity the weight of the rule acts through the pivot and has zero perpendicular distance, so it produces no moment and does not appear in the balance equation.

State the equation used in the unknown-mass balance, defining each symbol and its unit.

The balance condition is m₁d₁ = m₂d₂, where m₁ and m₂ are the two masses in kilograms and d₁ and d₂ are their perpendicular distances from the pivot in metres.

Explain why the masses in a metre-rule balance need not be converted into weights.

Both moments are a mass times g times a distance, so the factor g appears on both sides of the equation and cancels, leaving the masses and distances.

A metre rule balanced at its centre carries an unknown mass 20.0 cm from the pivot and a 100 g mass 32.0 cm from the pivot on the other side. Calculate the unknown mass.

Balancing gives m × 20.0 = 100 × 32.0, so m = 3200 ÷ 20.0 = 160 g; checking back, 160 × 20.0 = 3200 = 100 × 32.0, so the moments balance.

CCEA A-Level Physics 1210

Physical Quantities, Scalars and Vectors

Topic overview Physical quantities and SI base units Prefixes and standard form Scalars and vectors Resolving vectors and finding resultants Coplanar forces in equilibrium

Moments and Equilibrium

Topic overview Moment of a force Centre of gravity The principle of moments Determining unknown masses using moments

Linear Motion and Dynamics

Topic overview Displacement, velocity and acceleration The equations of motion for uniform acceleration Measuring the acceleration of free fall with light gates Displacement–time and velocity–time graphs Projectile motion Projectile calculations

Newton's Laws of Motion

Topic overview Newton's first law and inertia Newton's second law and F = ma Applying Newton's laws to simple situations Newton's third law Friction as a force opposing motion

Linear Momentum and Impulse

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Work, Energy and Power

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Current, Charge and Potential Difference

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Resistance and Resistivity

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Internal Resistance and Potential Dividers

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Waves

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Refraction

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Quantum Physics

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Wave-Particle Duality

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Astronomy

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Practical Skills and Data Analysis

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Deformation of Solids

Topic overview Hooke's law and the spring constant Elastic and plastic deformation and the elastic limit Stress, strain and the Young modulus Measuring the Young modulus of a wire Strain energy and force–extension graphs Choosing materials: strength, stiffness and cost

Thermal Physics

Topic overview Boyle's law, Charles's law and the pressure law The ideal gas equation pV = nRT The Avogadro constant, the mole and pV = NkT Internal energy and the distribution of molecular energy The kinetic theory equation pV = ⅓Nm⟨c²⟩Average molecular kinetic energy and temperature Absolute zero and the kelvin scale Specific heat capacity and the electrical method

Uniform Circular Motion

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Simple Harmonic Motion

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The Nucleus and Radioactive Decay

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Nuclear Energy, Fission and Fusion

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Gravitational Fields

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Electric Fields

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Magnetic Fields

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Fundamental Particles

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Moments and Equilibrium

Determining unknown masses using moments

3 things to know
Unit AS 1

Overview

Recall questions on balancing a metre rule to find an unknown mass, why it is first balanced at its centre of gravity, why masses need not be converted to weights, finding the mass of the rule itself by moving the pivot, the main sources of uncertainty, and why several known masses are used.

What you need to know

Answer each one before you reveal it.

Your recallNot started

Key term

Describe how the principle of moments is used to determine an unknown mass with a metre rule.

Key term

Describe how the mass of a uniform metre rule is determined using the principle of moments.

Key term

Explain why the pivot must be moved away from the centre when the mass of a uniform rule is being found.

Questions from this subtopic

Each one with the model answer underneath.

1
Describe how the principle of moments is used to determine an unknown mass with a metre rule.
Model answer
The rule is balanced on a knife edge at its centre of gravity, the unknown mass is hung a measured distance on one side and a known mass is hung on the other, and the known mass is moved until the rule balances again, so that the unknown mass times its distance equals the known mass times its distance.
2
Explain why the metre rule is first balanced at its centre of gravity before any masses are hung on it.
Model answer
With the pivot at the centre of gravity the weight of the rule acts through the pivot and has zero perpendicular distance, so it produces no moment and does not appear in the balance equation.
3
State the equation used in the unknown-mass balance, defining each symbol and its unit.
Model answer
The balance condition is m₁d₁ = m₂d₂, where m₁ and m₂ are the two masses in kilograms and d₁ and d₂ are their perpendicular distances from the pivot in metres.
4
Explain why the masses in a metre-rule balance need not be converted into weights.
Model answer
Both moments are a mass times g times a distance, so the factor g appears on both sides of the equation and cancels, leaving the masses and distances.
5
A metre rule balanced at its centre carries an unknown mass 20.0 cm from the pivot and a 100 g mass 32.0 cm from the pivot on the other side. Calculate the unknown mass.
Model answer
Balancing gives m × 20.0 = 100 × 32.0, so m = 3200 ÷ 20.0 = 160 g; checking back, 160 × 20.0 = 3200 = 100 × 32.0, so the moments balance.
6
Describe how the mass of a uniform metre rule is determined using the principle of moments.
Model answer
The rule is pivoted at a knife edge away from its 50.0 cm mark so that it no longer balances, and a known mass is hung on the shorter side and moved until the rule is horizontal; taking moments about the knife edge, the known weight times its distance equals the rule's weight times the distance from the knife edge to the 50.0 cm mark.

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