Topic 1 of 4
Units, dimensions and estimates
A physical quantity describes something that can be measured or calculated. State the value and its unit where applicable, then check whether its size makes sense.
In l = 2.0 m, l is a quantity symbol for length, 2.0 is the numerical value, and m is the unit symbol for metre. Writing the same length as 200 cm changes the numerical value and unit together; it does not change the length.
Some quantities are dimensionless. For example, the ratio of two lengths, 6.0 m / 2.0 m = 3.0, has no remaining unit because the metres cancel. Still explain what the ratio compares.
Useful starting skills are powers of ten, rearranging an equation and working with fractions. For example, 10-3 means 1/1000, and dividing by 10-3 is the same as multiplying by 1000.
The six base quantities required here
Derived units are built from base units. Learn both the quantity and its unit, including the case of each symbol. These are the six base quantities used in this course's introductory unit work.
| Quantity | Unit name | Symbol |
|---|---|---|
| Mass | kilogram | kg |
| Length | metre | m |
| Time | second | s |
| Electric current | ampere | A |
| Thermodynamic temperature | kelvin | K |
| Amount of substance | mole | mol |
The SI base unit of mass is kg, not g. Kelvin is K, with no degree sign. A mole, mol, is a unit of amount of substance; it is not a unit of mass.
Thermodynamic temperature is commonly written T and measured in K. Celsius temperature may be written θ or T and measured in °C; read the stated convention and unit. The scales have different zero points: T/K = θ/°C + 273.15, so 20 °C = 293.15 K.
A change of 1 °C has the same size as a change of 1 K. For example, warming from 20 to 50 °C is a 30 °C or 30 K interval. Ratios of Celsius readings do not give ratios of thermodynamic temperatures or energies: 40 °C is not twice 20 °C on the kelvin scale.
A prefix multiplies the unit
| Prefix | Symbol | Factor |
|---|---|---|
| pico | p | 10-12 |
| nano | n | 10-9 |
| micro | µ | 10-6 |
| milli | m | 10-3 |
| centi | c | 10-2 |
| deci | d | 10-1 |
| kilo | k | 103 |
| mega | M | 106 |
| giga | G | 109 |
| tera | T | 1012 |
Case matters: m means milli when used as a prefix, while M means mega. For example, 1 mA = 10-3 A but 1 MA = 106 A. In 1 mm, the first m is the prefix milli and the second is the unit metre.
Worked conversions
Apply the factor to the whole unit
A current: 250 µA = 250 × 10-6 A = 2.50 × 10-4 A.
An area: 1 cm = 10-2 m, so 1 cm2 = (10-2 m)2 = 10-4 m2. Therefore 4.0 cm2 = 4.0 × 10-4 m2.
A density: 1 g = 10-3 kg and 1 cm3 = 10-6 m3. Hence 2.0 g/cm3 = (2.0 × 10-3)/(10-6) kg/m3 = 2.0 × 103 kg/m3.
In a powered unit, raise the conversion factor to that power as well. Converting cm2 to m2 uses 10-4, not 10-2.
Volume is usually written V, or sometimes v under a stated convention, and its SI unit is m3. Area A has unit m2. Use volume in density ρ = m/V, but the relevant contact area in pressure p = F/A; a symbol such as V must be interpreted in its stated context.
Optional check An area is 12 mm^2. What is its value in m^2?
Derive a unit from its physical relationship
Replace the quantities in an equation by their units, then simplify. For example, acceleration is change in velocity per unit time, so its unit is (m/s)/s = m s-2. Negative powers represent division: s-2 means 1/s2.
- Force: the newton, N
- For mass m with acceleration a, F = ma. Therefore N = kg m s-2.
- Work and energy: the joule, J
- For a constant force parallel to a displacement, work = force × displacement. Therefore J = N m = kg m2 s-2.
- Power: the watt, W
- Power is energy transferred per unit time. Therefore W = J/s = kg m2 s-3.
- Charge: the coulomb, C
- A constant current I transfers charge Q = It in time t. Therefore C = A s.
- Potential difference: the volt, V
- Potential difference is energy transferred per unit charge: V = E/Q. Therefore V = J/C = kg m2 s-3 A-1.
- Pressure: the pascal, Pa
- Pressure is normal force per unit area, p = F/A. Therefore Pa = N/m2 = kg m-1 s-2.
Symbols need context. In p = F/A, A is the quantity symbol for area; after a numerical current, A is the unit ampere. C after a charge value means coulomb. A quantity symbol used in another equation need not have that same meaning.
Pressure and units
A force spread over an area
A normal force of 20 N acts uniformly over 4.0 × 10-3 m2. The pressure is p = 20/(4.0 × 10-3) = 5.0 × 103 Pa.
The unit standard atmosphere, atm, is defined by 1 atm = 101325 Pa. Thus 0.50 atm is about 5.1 × 104 Pa. Actual atmospheric pressure varies with conditions and altitude; the unit atm has a fixed definition.
Homogeneity checks every term
An equation is dimensionally homogeneous when both sides have the same dimensions. Terms added or subtracted must also have the same dimensions. Expressing their units in SI base units makes the comparison clear.
Checking an equation
Displacement, velocity and acceleration
Consider s = ut + ½at2, where s denotes displacement, u initial velocity, a acceleration and t time. Here the quantity symbol s must not be confused with the unit symbol s for second.
- Displacement has units m.
- ut has units (m s-1)s = m.
- ½at2 has units (m s-2)s2 = m; ½ is dimensionless.
Each term has length dimensions. In the incorrect expression s = ut + at, however, at has units m s-1. Adding that speed term to the length ut is not homogeneous.
A successful unit check does not prove an equation. The expression s = ut + at2 is also homogeneous. Dimensions cannot determine the factor ½ or establish the physical assumptions behind a relationship.
Likewise, when a trigonometric, exponential or logarithmic function appears in a physical equation, its argument must be dimensionless. Units provide a constraint on the expression, not a derivation of the whole model.
Optional check In the proposed equation s = ut + at^2, s is displacement, u is velocity, a is acceleration and t is time. What does a unit check establish?
Estimate with stated assumptions
Choose a plausible reference, state your assumptions and keep only justified precision. An estimate is useful when it can distinguish a sensible answer from one that is many powers of ten too large or too small.
- Floor area: model a roughly rectangular classroom as 8 m by 6 m. Its area is about 50 m2. The multiplication gives 48, but assumed dimensions do not justify reporting 48.000 m2.
- Mass: for an ordinary adult, a rough assumed range of 50 to 90 kg may be appropriate, with 60 kg as one working estimate. The chosen person's size matters; this is not a universal measured mass.
- Walking time: assume a 500 m route and a walking speed of 1.5 m/s. Then time is roughly 500/1.5 ≈ 300 s, or about 5 min. Stops or a slower pace would increase it.
- Small scales: a supplied microscopic length of 2 µm is 2 × 10-6 m; its scale is micrometres, not millimetres. A supplied current of 0.20 mA is 2.0 × 10-4 A, a few ten-thousandths of an ampere.
Estimate from a simple shape
Volume, density and contact pressure
Model a rigid rectangular block as roughly 10 cm long, 5 cm wide and 2 cm thick, with mass about 0.2 kg. These are rough assumptions, not precise measured dimensions.
= 1 × 10-4 m3
ρ = m/V ≈ 0.2/(1 × 10-4)
= 2 × 103 kg/m3
The volume is about one tenth of a litre. The density is on the scale of 103 kg/m3; the approximate inputs do not justify several decimal places.
Let the block rest on its 10 cm by 5 cm face on a horizontal table. Assume weight and support are its only vertical forces. With supplied g about 10 N/kg, the block presses down on the table with force about mg = 0.2 × 10 = 2 N.
= 5 × 10-3 m2
Average contact pressure p = F/A
≈ 2/(5 × 10-3) = 400 Pa
This is a few hundred pascals, between 102 and 103 Pa. The estimate uses the whole flat face as the contact area; local pressure need not be uniform. A smaller contact face gives a greater average pressure at the same force.
Keep the conversions distinct: 1 cm3 = 10-6 m3, while 1 cm2 = 10-4 m2. Volume is used for density; the selected contact area is used for pressure.
Select the geometric quantity
Coating area is different from volume or contact area
A thin coating over an object's entire outside uses its total surface area. Density uses its volume. Identify which physical quantity is needed before choosing a formula; ignore coating thickness in these geometric models.
A triangular sheet needs a perpendicular height
A thin flat sheet has an isosceles triangular face with supplied base 6.00 cm and equal sides 5.00 cm. Symmetry divides it into two right triangles, each with base 3.00 cm. Pythagoras gives the perpendicular height h:
Area of each right half = (1/2)(3.00)(4.00)
= 6.00 cm2
Area of the whole isosceles face = (1/2)(6.00)(4.00)
= 12.0 cm2
This is the area of one physical face, using perpendicular height rather than a sloping side. Covering both broad faces requires 24.0 cm2 if the thin edge area is neglected.
A rectangular block has three pairs of faces
For perpendicular edge lengths l, w and h, V = lwh and total surface area S = 2(lw + lh + wh). The rough 10 cm by 5 cm by 2 cm block above would need coating over:
= 160 cm2 = 0.016 m2
This includes all six faces. It is different from the single 50 cm2 contact face used for pressure and the 100 cm3 volume used for density.
A cylinder has a curved side and two circular ends
Model a solid right circular cylinder with supplied radius r = 1.00 cm and length L = 10.0 cm. Its circular end area is πr2. Unrolling the curved side gives a rectangle of width 2πr, the circumference, and length L.
≈ 31.4 cm3
S = 2πrL + 2πr2 = 22π cm2
≈ 69.1 cm2
Coating the side and both ends requires about 69.1 cm2. If this cylinder has supplied mass 50.0 g, its density is 50.0/(10π) = 1.59 g/cm3, about 1.59 × 103 kg/m3. A measured diameter d must first be halved: r = d/2, so V = πd2L/4.
A sphere uses radius, not diameter
Model a solid spherical bead with supplied diameter 2.00 cm, so r = 1.00 cm. Its whole outside and occupied volume are:
≈ 12.6 cm2
V = (4/3)πr3 = (4/3)π cm3
≈ 4.19 cm3
A thin coating covers about 12.6 cm2, not the projected circular area πr2 = 3.14 cm2. With supplied bead mass 8.40 g, density is 8.40/[(4/3)π] = 2.01 g/cm3. These formulas model a complete sphere, not a hollow shell or an irregular bead.
After a calculation, compare the result with the assumptions and units. If the classroom estimate becomes 500000 m2, revisit the conversion before interpreting the physics.