Full chapter
Quantities and Measurement
All 4 topics and the revision summary on one page.
01
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.
02
Errors, accuracy and precision
Repeated readings can agree closely and still be displaced from the correct value. Identify what caused a limitation before choosing how to improve the measurement.
- Error
- The difference between a measured value and a reference or true value. The true value, and hence the exact error, is often unknown.
- Uncertainty
- The doubt associated with a result, often expressed as an estimated range around it. An uncertainty is not automatically a known correction.
- Accuracy
- Closeness to the true or accepted reference value.
- Precision
- Closeness of agreement between repeated measurements under the stated conditions. A narrow scatter indicates greater precision.
Compare supplied repeated length readings
Each dot is one reading. All three rows use the same horizontal scale. The dashed reference is 20.0 mm; the row height only separates the sets.
Sets 1 and 2 have equally small scatter here, but set 2 is displaced from the reference. Set 3 has a mean close to the reference with less precise individual readings. Averaging does not automatically remove a persistent bias.
The narrow displaced set has mean 20.6 mm, compared with the reference 20.0 mm. Repeating this biased procedure many times could produce a very consistent mean near the wrong value. More display digits would not repair the cause.
Systematic effects follow a pattern
A systematic error shifts readings in a consistent way under the stated conditions. A zero offset adds a fixed amount; a calibration scale error may instead multiply readings by a factor.
Correcting a zero offset
Use the sign of the empty reading
A balance reads +0.20 g with no load and 24.80 g with an object. If the zero offset remains unchanged, corrected mass = displayed mass - offset = 24.80 - 0.20 = 24.60 g.
This removes the known additive offset. Resolution, random fluctuations and other calibration effects can still contribute uncertainty.
A ruler that consistently indicates lengths 2% too high has a scale-factor error. It would indicate 51 cm for a true 50 cm length and 102 cm for a true 100 cm length. Subtracting one fixed number of centimetres cannot correct both. Under this supplied model, divide each indicated length by 1.02.
Random variation produces scatter
Random effects make repeated readings vary. For suitable independent repeats, using a mean reduces the effect of random variation on the estimated value. Keep the observed spread as evidence of uncertainty; averaging does not make it vanish or remove a persistent calibration error.
Read and calculate a mean
Two symbols for the same arithmetic mean
The arithmetic mean of readings x can be written <x> or , pronounced x bar. The symbol Σ means sum the specified readings. Here n counts readings and is dimensionless:
<x> = = (Σxi)/n
For the supplied 19.8, 19.9, 20.0, 20.1 and 20.2 mm set, the sum is 100.0 mm and n = 5. Thus <x> = 100.0/5 = 20.0 mm. It retains the readings' unit. This mean does not erase their spread or establish that every reading is exactly 20.0 mm.
In this notation <x> encloses a quantity to indicate its mean. A statement such as x < y instead uses < to mean less than.
Describe the actual mechanism. A stopwatch user who consistently starts late, with no matching delay when stopping, biases the interval low. Variation in reaction time from trial to trial also causes random scatter. The word "human" alone does not tell you which kind of error is present.
Repeating measurements is helpful for random variation. A stable zero error needs a zero check and correction; a wrong scale factor needs calibration. An improvement should address the identified effect.
Optional check A balance displays -0.30 g with no load and 18.20 g with an object. The offset stays unchanged. Which conclusion is justified?
Choose range, resolution and a suitable method
Resolution is the smallest change an instrument indicates. It is one consideration when judging uncertainty, but it is not automatically the total uncertainty of a reading or procedure. Use a supplied uncertainty or justify an estimate for the actual method.
Instrument choice
Measure a rod about 120 mm long
The target is an uncertainty no greater than 0.1 mm. Use the following supplied specifications.
Choose the calipers, check the zero and hold the rod square to the jaws without compressing it. The 0.01 mm display step does not justify replacing the supplied ±0.05 mm uncertainty with ±0.01 mm.
There is no universal rule that every measurement has uncertainty equal to half its smallest division. Reading two ends, positioning the object, reaction time and calibration can change what is justified.
For a repeated motion timed by hand, timing many complete cycles can reduce the fractional effect of a similar start/stop timing uncertainty. Count accurately and use the same reference point and direction. A longer interval does not correct an incorrect clock calibration.
When a result is calculated from measurements, carry the relevant uncertainties into the derived result.
03
Uncertainty in calculated results
A calculated value cannot be more certain than its inputs justify. The operation in the equation determines how the input uncertainties contribute.
Write a measurement as x ± Δx. The absolute uncertainty Δx is nonnegative and has the same unit as x. For a value well away from zero:
Fractional uncertainty = Δx / |x|
Percentage uncertainty = (Δx / |x|) × 100%
For example, (5.0 ± 0.1) cm has fractional uncertainty 0.1/5.0 = 0.02 and percentage uncertainty 2%. These estimates describe the stated measurement limits; they do not, by themselves, specify a statistical confidence level.
The condition that a fractional uncertainty is small may be written Δx/|x| ≪ 1, meaning much less than 1. Here 0.02 ≪ 1; equivalently, the ratio |x|/Δx = 50 ≫ 1 is much greater than 1. These symbols express a scale comparison, not a universal numerical cut-off for every approximation.
Lowercase δx often denotes a small change. If a length increases from 2.000 m to 2.003 m, δx = +0.003 m and its percentage change is (0.003/2.000) × 100% = +0.15%. A decrease gives a negative change.
Read the stated convention: here Δx in x ± Δx is a nonnegative absolute uncertainty, whereas in a motion equation Δx commonly means the signed final-minus-initial change. Some texts use δx for uncertainty too. The Greek letter alone does not determine the meaning or sign.
Select the rule from the operation
- Addition or subtraction: add absolute contributions
- For q = x + y or q = x - y, a conservative estimate is Δq = Δx + Δy, for separately bounded inputs with no shared effect being cancelled.
- Multiplication or division: add fractional contributions
- For q = xy or q = x/y, with small relative uncertainties, Δq/|q| ≈ Δx/|x| + Δy/|y|. Percentage contributions can be added in the same way.
- A power: multiply by the absolute exponent
- For q = xn, the small-uncertainty estimate is Δq/|q| ≈ |n| Δx/|x|. A squared diameter contributes twice its fractional uncertainty, even if it is in the denominator.
- An exact factor: no extra measured contribution
- For q = kx with exact k, the fractional uncertainty is unchanged. The absolute uncertainty scales by |k|. Constants such as 4 and π in a supplied geometric formula do not add measurement uncertainty.
Worked difference
Subtract values, add their uncertainty contributions
Two separately bounded length readings are (31.6 ± 0.1) cm and (12.4 ± 0.1) cm.
Difference = 31.6 - 12.4 = 19.2 cm.
Absolute uncertainty = 0.1 + 0.1 = 0.2 cm.
The result is (19.2 ± 0.2) cm. The greatest possible difference uses the first reading high and the second low: 31.7 - 12.3 = 19.4 cm. The smallest is 31.5 - 12.5 = 19.0 cm.
Check for shared effects before applying a rule. If the same fixed zero offset z affects both readings on one unchanged scale, (x + z) - (y + z) = x - y. That offset cancels. It should not be counted twice as unrelated uncertainty, although other reading uncertainties may remain.
Optional check Two separately bounded readings are (43.2 +/- 0.2) cm and (11.8 +/- 0.1) cm. With no shared offset to cancel, what is their difference and conservative uncertainty?
Worked quotient
Speed from a distance and a time
A distance L = (10.00 ± 0.02) m is travelled in t = (2.50 ± 0.05) s. Find the average speed v = L/t and its estimated uncertainty.
- Calculate the value: v = 10.00/2.50 = 4.00 m/s.
- Add fractional contributions: Δv/v ≈ 0.02/10.00 + 0.05/2.50 = 0.002 + 0.020 = 0.022.
- Return to an absolute uncertainty: Δv ≈ 4.00 × 0.022 = 0.088 m/s.
- Report sensibly: approximately (4.00 ± 0.09) m/s.
The time contributes 2.0%, compared with the distance's 0.2%. Reducing the timing uncertainty is the more effective priority in this example.
Numerical substitution checks the bounds
For positive v = L/t, increasing L increases v while increasing t decreases it. Choose the input extremes accordingly:
vmax = 10.02/2.45 = 4.0898 m/s
vmin = 9.98/2.55 = 3.9137 m/s
The bounds are close to 4.00 ± 0.09 m/s, but not exactly symmetric about 4.00. This supports the small-uncertainty approximation here. Putting every input at its maximum would not give the maximum of a quotient.
If a denominator's possible range includes zero, or relative uncertainties are large, the simple fractional rule may be unsuitable. Inspect how the supplied expression changes over the allowed input ranges and use numerical bounds where appropriate. For a function that turns within the range, its maximum or minimum need not occur at an endpoint.
Worked power and quotient
Which cylinder measurement matters most?
For a uniform cylinder, volume = πd2L/4 and density is mass divided by volume, so use the supplied model ρ = 4m/(πd2L).
| Quantity | Value and absolute uncertainty |
|---|---|
| Mass m | (0.0500 ± 0.0001) kg |
| Diameter d | (0.0200 ± 0.0002) m |
| Length L | (0.100 ± 0.001) m |
Using the central values gives ρ = 1591.55 kg/m3 before rounding.
The fractional contributions are:
- Mass: 0.0001/0.0500 = 0.002.
- Diameter squared: 2(0.0002/0.0200) = 0.020.
- Length: 0.001/0.100 = 0.010.
Total fractional estimate = 0.032, or 3.2%. The absolute estimate is 1591.55 × 0.032 = 50.93 kg/m3. A suitable reported result is (1.59 ± 0.05) × 103 kg/m3.
The diameter contributes 2.0%, the length 1.0% and the mass 0.2%. Halving the diameter's percentage uncertainty reduces the total more than halving either of the other percentage uncertainties. Improving an already precise mass reading has little effect here.
Keep guard digits during calculation. Round the final uncertainty sensibly, then give the value to a corresponding place. The appropriate precision depends on the evidence; one rule about decimal places does not apply to every raw instrument reading.
Optional check For density rho = 4m/(pi d^2 L), the percentage uncertainties in m, d and L are 0.2%, 1.0% and 1.0%. Which single improvement most reduces the estimated density uncertainty?
04
Vectors and perpendicular components
A vector includes direction as well as magnitude. Account for both when combining forces, displacements or velocities.
| Scalars: magnitude only | Vectors: magnitude and direction |
|---|---|
| Mass, temperature, energy, time, distance and speed | Displacement, velocity, acceleration and force |
A temperature of -5 °C is still a scalar: it has no spatial direction. A signed coordinate is not, just because it has a minus sign, a complete vector description. For a one-dimensional velocity, a sign can encode direction once the positive axis has been stated.
Represent a vector by an arrow: length represents magnitude, and the arrowhead shows direction. Vectors in the same plane are coplanar. Use fixed axes when calculating their components; rotating the axes changes the component values, not the physical vector.
Add vectors head to tail
Draw the first vector, then place the tail of the second at the head of the first. Preserve the second vector's length and direction. The resultant points from the first tail to the final head.
Worked addition
East and north components
Take east as +x and north as +y. Let A = (3, 4) m and B = (2, -1) m. A points 3 m east and 4 m north; B points 2 m east and 1 m south.
Add corresponding components: A + B = (5, 3) m.
Magnitude = √(52 + 32) = √34 = 5.83 m.
Direction = tan-1(3/5) = 31.0° north of east.
A + B: join head to tail
A = (3, 4) m and B = (2, -1) m use east and north as positive. Move B without turning or stretching it, putting its tail at the tip of A. The purple resultant goes from the original start to the final tip.
A + B = (5, 3) m. Its magnitude is 5.83 m and its direction is 31.0° north of east. The component signs and the arrow's quadrant agree.
A paper scale drawing can also give the magnitude and direction: state a scale, draw the arrows head to tail, then measure the resultant and convert its length back. A resized screen diagram is not a physical centimetre ruler.
Subtract by reversing the second vector
A - B = A + (-B). The vector -B has the same magnitude as B and the opposite direction. Reverse B, then perform the same head-to-tail construction.
Worked subtraction
Use the signs of both components
For the same A and B, -B = (-2, +1) m. Therefore:
A - B = (3 - 2, 4 - (-1)) = (1, 5) m.
Magnitude = √(12 + 52) = √26 = 5.10 m.
Direction = tan-1(5/1) = 78.7° north of east.
A - B: add the reversed B
A = (3, 4) m and B = (2, -1) m use east and north as positive. Reverse B without changing its length, then put the tail of -B at the tip of A. The purple resultant goes from the original start to the final tip.
A - B = (1, 5) m. Its magnitude is 5.10 m and its direction is 78.7° north of east. The component signs and the arrow's quadrant agree.
Subtracting the magnitudes of A and B would ignore their directions. Coplanar vector subtraction is performed using the arrows or their signed components.
Resolve into perpendicular components
Perpendicular components are an equivalent way of representing one vector. The original vector is the hypotenuse of a right triangle. The component adjacent to the stated angle uses cosine; the opposite component uses sine.
Worked resolution
A 10.0 N force at 30.0°
Take right as +x and up as +y. The angle is measured above the positive horizontal.
Fx = 10.0 cos 30.0° = +8.66 N.
Fy = 10.0 sin 30.0° = +5.00 N.
The same force is at 60.0° to the positive vertical. Using that angle, Fx = 10.0 sin 60.0° and Fy = 10.0 cos 60.0°, giving the same components.
For a different 10.0 N force at 150° from +x, the arrow points left and up. Its components are Fx = -8.66 N and Fy = +5.00 N.
Resolve one force into perpendicular components
Purple is the original 10.0 N force. Green arrows represent its horizontal and vertical components placed head to tail. They are an equivalent description of that force, not two extra forces acting alongside it.
One direction, two angle references
Using the horizontal reference: Fx = 10.0 cos 30° and Fy = 10.0 sin 30°. From the vertical reference, use Fx = 10.0 sin 60° and Fy = 10.0 cos 60°. The vector and its components stay the same.
The identity sin2θ + cos2θ = 1 gives Fx2 + Fy2 = F2(cos2θ + sin2θ) = F2. Here (8.66 N)2 + (5.00 N)2 is approximately 100 N2, recovering the original 10.0 N magnitude.
The horizontal component is negative
The direction is 150° from +x, or 30° above -x. The components are (-8.66, +5.00) N: left and up. An inverse-tangent value must be interpreted in the correct quadrant.
Use the geometry of the component triangle
Angles and similar triangles
The interior angles of a triangle sum to 180°. The right triangle for the force at 30° therefore has third angle 180° - 90° - 30° = 60°. This is why the same force can be described as 30° above horizontal or 60° from vertical.
Triangles with equal corresponding angles are similar: corresponding side lengths have the same scale factor. A 5.00 N force at that same 30° has a component triangle similar to the 10.0 N triangle. The scale factor is 5.00/10.0 = 0.500, so its components are 0.500 × 8.66 = 4.33 N horizontally and 0.500 × 5.00 = 2.50 N vertically.
Both components scale together and the angle is unchanged. Similarity cannot be assumed for two triangles merely because they are both right-angled; their remaining corresponding angles must agree too.
The components replace the original force in the calculation. They are not two extra forces to add alongside it on the same free-body diagram.
The identity sin2θ + cos2θ = 1 explains why recombining the two perpendicular components recovers the original magnitude, as shown in the diagram.
To reconstruct a vector from perpendicular components, use magnitude √(Fx2 + Fy2) and determine the quadrant from the signs. An inverse tangent alone can return an angle in the wrong quadrant. For (-8.66, +5.00) N, the direction is 30° north of west, or 150° anticlockwise from +x.
A velocity change is a vector difference
Worked application
Turning from east to north
An initial velocity is 8.0 m/s east and a final velocity is 6.0 m/s north. Take east and north as positive.
Δv = vfinal - vinitial
= (0, 6.0) - (8.0, 0)
= (-8.0, +6.0) m/s.
Magnitude = √(8.02 + 6.02) = 10.0 m/s.
Direction = tan-1(6.0/8.0) = 36.9° north of west.
Velocity change runs from the initial tip to the final tip
Draw the initial and final velocities from the same origin. The purple arrow from the initial tip to the final tip is the change. It points 8.0 m/s west and 6.0 m/s north.
Change in velocity = (-8.0, +6.0) m/s, with magnitude 10.0 m/s and direction 36.9° north of west. A time interval would also be needed to calculate an average acceleration.
The speed decreases by 2.0 m/s, but the magnitude of the velocity change is 10.0 m/s because the direction changes too. The two velocity values alone do not establish an acceleration or show that it is constant; time information and a motion model are needed.
Optional check A velocity changes from 6.0 m/s east to 8.0 m/s south. Taking east and north as positive, what is the change in velocity?
Revision summary
State a result's value, its unit where applicable, and justified precision. Explain dimensionless ratios, and include direction for a vector. Use the linked explanations when you need to reconstruct a method.
Base units and prefixes
| 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 |
| 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 |
Preserve case, including milli m and mega M. Square an area conversion factor and cube a volume conversion factor. A quantity symbol and a unit symbol have different roles.
Volume V, or v under a stated convention, has SI unit m3. Use volume for density, the selected contact face for average pressure, and total outside area for a complete coating. Count all six block faces, both cylinder ends when included, and the whole spherical surface; do not substitute a sphere's projected circle for its surface area.
Temperature symbols θ or T follow the stated convention. Thermodynamic temperature uses K; Celsius temperature uses °C. T/K = θ/°C + 273.15, and a temperature interval has equal numerical values in K and °C. Do not use ratios of Celsius readings as thermodynamic-temperature ratios.
- Force
- N = kg m s-2
- Energy
- J = N m = kg m2 s-2
- Power
- W = J/s = kg m2 s-3
- Charge
- C = A s
- Potential difference
- V = J/C = kg m2 s-3 A-1
- Pressure
- Pa = N/m2 = kg m-1 s-2. The fixed unit definition is 1 atm = 101325 Pa; actual atmospheric pressure varies.
Homogeneity: compare every added term and both sides using base units. A unit check can reject an equation, but cannot prove it or determine a dimensionless numerical factor. For an estimate, state plausible assumptions and avoid false precision.
Errors and measurement choices
- Accuracy: closeness to a reference. Precision: agreement between repeats.
- Random variation: produces scatter; suitable repetition and a mean can reduce its effect on the estimate.
- Systematic effects: persist in the mean. Correct a stable zero offset by subtracting its signed value; check calibration for a scale-factor error.
- Resolution: smallest indicated change, not automatically the total measurement uncertainty.
- Choose enough range and suitable uncertainty, then control alignment, zero, loading and timing as appropriate. Match each improvement to the actual limitation.
The arithmetic mean is <x> = = (Σxi)/n: add the specified readings and divide by their count. The mean has the same unit as the readings.
Derived uncertainty
- Absolute, fractional, percentage
- Δx has the unit of x and is nonnegative. Fractional = Δx/|x|; percentage = fractional × 100%.
- Sum or difference
- Add the relevant absolute uncertainty contributions conservatively. A common fixed offset can cancel in a difference.
- Product or quotient
- For small relative uncertainties, add fractional or percentage contributions.
- Power xn
- Multiply its fractional contribution by |n|. An exact constant adds no measured uncertainty.
- Numerical limits
- Choose input combinations from the expression's behaviour. For positive L/t, maximum uses high L and low t; minimum uses low L and high t.
Retain guard digits, then report an uncertainty and value with compatible precision. Large relative uncertainties or a denominator range containing zero require more care. These uncertainty estimates are not automatically statistical confidence intervals.
≪ means much less than; ≫ means much greater than. A small-uncertainty approximation needs a suitable scale comparison such as Δx/|x| ≪ 1. Lowercase δx may denote a small signed change, while this section uses Δx for nonnegative uncertainty; follow the stated convention.
Vectors and components
- Scalars include mass, temperature, energy and speed. Vectors include displacement, velocity, acceleration and force. A minus sign alone does not define a vector.
- Addition: preserve arrow lengths and directions, place them head to tail, and draw the resultant from the first tail to the final tip.
- Subtraction: A - B = A + (-B). Reverse B, then add; do not merely subtract magnitudes.
- Resolution: adjacent component uses cosine, opposite component uses sine. Identify the reference angle and choose signs from the axes.
- Reconstruction: magnitude = √(x2 + y2); determine direction with the correct quadrant.
- Velocity change: Δv = vfinal - vinitial. Its magnitude need not equal the change in speed.
Components are an equivalent representation of the original vector. Do not count them as extra forces alongside it.
A triangle's interior angles sum to 180°. Similar triangles have equal corresponding angles and proportional corresponding sides, so scaling a vector at a fixed angle scales both of its perpendicular components by the same factor.
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