Mechanics

Quantities & Units in Mechanics

Year 12 · Year 13

  • By the end of this lesson students will be able to recall and use the SI base units for mass, length, and time.
  • By the end of this lesson students will be able to understand and use derived SI units for common mechanical quantities.
  • By the end of this lesson students will be able to check the consistency of units in equations using dimensional analysis.
  • By the end of this lesson students will be able to understand the concept of a mathematical model in mechanics.
  • By the end of this lesson students will be able to identify and state appropriate modelling assumptions for given physical situations.

Key concepts

SI Base Units

The International System of Units (SI) provides a coherent system of units for scientific and engineering measurements. In mechanics, the fundamental base units are for mass, length, and time. All other mechanical units can be derived from these three.

Derived SI Units

Derived units are combinations of base units. For example, velocity is a derived unit (length per time), and force is another (mass times acceleration). Understanding how derived units are formed from base units is crucial for dimensional analysis and ensuring consistency in calculations.

Dimensional Analysis

Dimensional analysis is the process of checking the consistency of units in an equation. Both sides of a valid physical equation must have the same base units. This technique can help identify errors in formulae or unit conversions.

Mathematical Modelling

A mathematical model is a simplified representation of a real-world situation, designed to make it amenable to mathematical analysis. Real-world problems are often too complex to solve directly, so assumptions are made to create a manageable model. The accuracy of the model depends on the appropriateness of the assumptions.

Common Modelling Assumptions

When creating a mathematical model in mechanics, specific assumptions are often made to simplify the problem. These assumptions are vital for solving the problem and must be clearly stated. Common assumptions include:

Particle

An object whose mass is concentrated at a single point. Its dimensions are negligible. This assumption is used when the size, shape, or rotational effects of an object are not relevant to the problem (e.g., a ball thrown through the air, where air resistance is considered but its spin is not).

Rigid Body

An object whose shape and size do not change under the action of applied forces. This assumption is used when the dimensions of an object are important (e.g., for moments or stability) but it does not deform (e.g., a ladder leaning against a wall).

Light String/Rod

A string or rod whose mass is negligible compared to the masses it connects or supports. This implies that the tension in a light string is uniform throughout its length (assuming no external forces acting along the string itself).

Inextensible String

A string whose length does not change under tension. This implies that objects connected by an inextensible string will have the same magnitude of acceleration along the line of the string.

Smooth Surface/Pulley

A surface or pulley where there is no friction. For a smooth surface, the reaction force is perpendicular to the surface. For a smooth pulley, the tension in the string passing over it is the same on both sides.

Rough Surface

A surface where friction is present. The frictional force opposes motion or the tendency of motion and its maximum value is given by F_max = μR, where μ is the coefficient of friction and R is the normal reaction force.

Uniform Rod/Lamina

A rod or lamina (flat plate) where the mass is evenly distributed throughout its length or area. This means its centre of mass is at its geometric centre.

Air Resistance Negligible

The resistive force exerted by the air on a moving object is ignored. This simplifies calculations significantly, as air resistance is often a complex function of velocity and shape.

Gravity Constant

The acceleration due to gravity (g) is assumed to be constant and uniform, typically taken as 9.8 m s⁻² (or sometimes 9.81 m s⁻² or 10 m s⁻² if specified) and acting vertically downwards.

Key facts to remember

  • 1The three SI base units in mechanics are the metre (m) for length, the kilogram (kg) for mass, and the second (s) for time.
  • 2Derived units are combinations of base units, e.g., velocity (m s⁻¹), acceleration (m s⁻²).
  • 3The Newton (N) is the SI unit of force, equivalent to kg m s⁻².
  • 4The Joule (J) is the SI unit of energy/work, equivalent to N m or kg m² s⁻².
  • 5The Watt (W) is the SI unit of power, equivalent to J s⁻¹ or kg m² s⁻³.
  • 6A 'particle' is a model where an object's mass is concentrated at a single point, ignoring its dimensions.
  • 7A 'light' string or rod has negligible mass.
  • 8An 'inextensible' string has a constant length, meaning connected objects have the same acceleration.
  • 9A 'smooth' surface or pulley implies no friction.

Worked examples

Example 1

Show that the equation for kinetic energy, E_k = 1/2 mv², is dimensionally consistent, where m is mass and v is velocity.

IIdentify the SI base units for each quantity:
IIMass (m): kilogram (kg)
IIIVelocity (v): metre per second (m s⁻¹)
IVKinetic Energy (E_k): Joule (J)
VExpress the Joule in terms of base units. We know Force (F) = mass × acceleration (ma), so 1 N = 1 kg m s⁻². Work/Energy (E) = Force × distance, so 1 J = 1 N m = 1 (kg m s⁻²) m = 1 kg m² s⁻².
VINow, substitute the base units into the right-hand side of the equation E_k = 1/2 mv²:
VIIUnits of (1/2 mv²) = Units of (m) × (Units of (v))²
VIII= kg × (m s⁻¹)²
9= kg × m² s⁻²
10Compare this with the base units for E_k (Joule): kg m² s⁻².
11Since the units on both sides of the equation are the same (kg m² s⁻²), the equation is dimensionally consistent. The numerical factor 1/2 has no units.

Answer

The equation E_k = 1/2 mv² is dimensionally consistent as both sides have the base units kg m² s⁻².

Dimensional analysis only checks for consistency of units, not the correctness of numerical constants.

Example 2

A student throws a tennis ball vertically upwards. List three suitable modelling assumptions that could be made to analyse its motion.

IConsider the object: A tennis ball. Its size and shape might affect air resistance, but for basic vertical motion, its internal structure or spin is often ignored.
IIConsider the forces: Gravity is always present. Air resistance is often a complex factor. The ball's dimensions are usually not critical for simple vertical motion.
IIIFormulate assumptions based on simplification:
IV1. The tennis ball can be modelled as a particle: This simplifies the problem by ignoring its size, shape, and any rotational effects, concentrating its mass at a single point.
V2. Air resistance is negligible: This simplifies the calculation of forces acting on the ball, as air resistance is often a complex function of velocity and shape. It allows us to consider only gravity.
VI3. The acceleration due to gravity is constant: We assume g = 9.8 m s⁻² (or 9.81 m s⁻² if specified) and acts uniformly downwards, ignoring variations with altitude or location.

Answer

Suitable modelling assumptions are: 1. The tennis ball is modelled as a particle. 2. Air resistance is negligible. 3. The acceleration due to gravity is constant (g = 9.8 m s⁻²).

Other valid assumptions could include 'the Earth is a fixed frame of reference' or 'no wind'.

Example 3

Determine the SI base units of pressure, given that Pressure = Force / Area.

IRecall the SI base units for Force and Area.
IIForce (F) has units of Newtons (N). From F = ma, 1 N = 1 kg m s⁻².
IIIArea (A) has units of square metres (m²).
IVSubstitute these into the given formula for Pressure:
VPressure = Force / Area
VIUnits of Pressure = (Units of Force) / (Units of Area)
VII= (kg m s⁻²) / (m²)
VIIISimplify the expression by cancelling out common terms:
9= kg m¹ s⁻² m⁻²
10= kg m⁻¹ s⁻²

Answer

The SI base units of pressure are kg m⁻¹ s⁻² (which is equivalent to the Pascal, Pa).

Pressure is often given in Pascals (Pa), where 1 Pa = 1 N m⁻². This example shows how to break down a derived unit into its fundamental base units.

Common mistakes

  • Using non-SI units (e.g., grams, centimetres, km/h) in calculations without converting them to kilograms, metres, and metres per second, respectively.
  • Confusing the modelling assumptions 'light' (negligible mass) with 'smooth' (no friction).
  • Forgetting to explicitly state all modelling assumptions made at the start of a problem, especially when asked to do so.
  • Incorrectly deriving the base units for complex quantities, often making errors with negative exponents or powers.
  • Assuming air resistance is always negligible without the problem stating it or without explicitly making and stating the assumption.

Exam tips

  • Always work in SI units (m, kg, s) unless the question specifically instructs otherwise. Convert all given values to SI units before starting calculations.
  • When asked to state modelling assumptions, be precise with your terminology (e.g., 'particle', 'light', 'inextensible', 'smooth', 'air resistance negligible').
  • Use dimensional analysis to check the consistency of your equations. If the units don't match on both sides, your equation is incorrect.
  • Practise deriving the base units for various physical quantities to become proficient in dimensional analysis.

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