Mp33–Mp35 - Mechanical quantities, SI units and dimensions

Mp33–Mp35 - Mechanical quantities, SI units and dimensions

Use the SI quantities in Mp33 to give mechanical measurements a common language, and build the velocity, acceleration, force and weight units in Mp34. Mp35 adds the newton metre for moments; unit checks will help you judge whether a proposed relationship can make sense.

Independent base quantities

One robot's controller reports an acceleration of 18000 cm min−218000\,\mathrm{cm\,min^{-2}}. Another reports 0.06 m s−20.06\,\mathrm{m\,s^{-2}}. Which acceleration has the larger magnitude? Those impressive-looking digits need translating before we can compare them.

Start with a simpler translation. A mass of 600 g600\,\mathrm g is 0.600 kg0.600\,\mathrm{kg}: the object has not become lighter when we change the unit.

You already know

A measurement has a number and a unit. A conversion replaces both consistently; it does not change the physical quantity.

Length, time and mass are the starting quantities for mechanics. Each describes a different property.

Fundamental SI quantities

Length is measured in metres (m)(\mathrm m), time in seconds (s)(\mathrm s), and mass in kilograms (kg)(\mathrm{kg}). These fundamental quantities are independent: none is made from the other two.

For example, two equal-mass trolleys need not have equal lengths or equal journey times. Knowing length and time does not determine mass. A physical model may connect quantities when more information is given, but a unit conversion cannot turn time into mass.

A unit is not the same as the quantity it measures. Mass is a property of a body; weight is a gravitational force on it. Kilograms measure mass, while newtons measure weight.

Key idea

Choose the quantity first, then its unit. The size of a written number alone does not compare two measurements.

Before comparing the robot controllers, check that you can separate a quantity from its unit.

Units built from relationships

The phrase “metres per second per second” tells you what acceleration measures: the change in velocity in each second.

A displacement change of 8 m8\,\mathrm m in 4 s4\,\mathrm s gives average velocity 2 m s−12\,\mathrm{m\,s^{-1}} along the chosen positive direction. Velocity has direction as well as magnitude; speed is the non-negative magnitude. Identical units do not make them identical quantities.

Derived quantities

Velocity is the rate of change of displacement. Acceleration is the rate of change of velocity. Force is a vector interaction that can change velocity; weight is the gravitational force acting on a body.

Acceleration divides a velocity change by time. Force has the base units of mass multiplied by acceleration; this is the unit connection behind the newton, before we use force equations in later lessons.

Mechanical SI units

velocity: m s−1acceleration: m s−2force and weight: N=kg m s−2\begin{aligned} \text{velocity: }&\mathrm{m\,s^{-1}}\\ \text{acceleration: }&\mathrm{m\,s^{-2}}\\ \text{force and weight: }&\mathrm N=\mathrm{kg\,m\,s^{-2}} \end{aligned}

A numerical answer needs the unit of the quantity asked for. Choose a positive direction before assigning a sign to a velocity or acceleration along a line.

Not in the formula sheet: learn it.

A small velocity change is enough to see where the second inverse-time factor comes from.

Worked example 1: acceleration from a velocity change

Question. A cart's velocity changes from 2 m s−12\,\mathrm{m\,s^{-1}} to 10 m s−110\,\mathrm{m\,s^{-1}} in 4 s4\,\mathrm s. Calculate its average acceleration.

Plan. Form final minus initial velocity, and divide by the elapsed time.

Δv=10−2=8 m s−1velocity changeaˉ=84=2 m s−2change per second\begin{aligned} \Delta v&=10-2=8\,\mathrm{m\,s^{-1}} &&\text{velocity change}\\ \bar a&=\frac{8}{4}=2\,\mathrm{m\,s^{-2}} &&\text{change per second} \end{aligned}

Why this step? The unit m s−1\mathrm{m\,s^{-1}} is itself divided by seconds, giving m s−2\mathrm{m\,s^{-2}}.

Answer. 2 m s−22\,\mathrm{m\,s^{-2}} in the positive direction.

Check. Multiplying 2 m s−22\,\mathrm{m\,s^{-2}} by 4 s4\,\mathrm s recovers the 8 m s−18\,\mathrm{m\,s^{-1}} change.

Now reverse only the order of the endpoint velocities. The unit remains an acceleration unit even when the answer is negative.

Worked example 2: a negative acceleration

Question. A cart's velocity changes from 10 m s−110\,\mathrm{m\,s^{-1}} to 2 m s−12\,\mathrm{m\,s^{-1}} in 4 s4\,\mathrm s. Calculate its average acceleration.

Plan. Keep the same positive direction and the same final-minus-initial order.

Δv=2−10=−8 m s−1velocity decreasesaˉ=−84=−2 m s−2divide by time\begin{aligned} \Delta v&=2-10=-8\,\mathrm{m\,s^{-1}} &&\text{velocity decreases}\\ \bar a&=\frac{-8}{4}=-2\,\mathrm{m\,s^{-2}} &&\text{divide by time} \end{aligned}

Why this step? A sign describes direction relative to your chosen axis; the acceleration's magnitude would instead be 2 m s−22\,\mathrm{m\,s^{-2}}.

Answer. −2 m s−2-2\,\mathrm{m\,s^{-2}}.

Check. The velocity decreases by 8 m s−18\,\mathrm{m\,s^{-1}}, consistent with the negative change.

Force and acceleration also need distinguishing. The product (3 kg)(2 m s−2)(3\,\mathrm{kg})(2\,\mathrm{m\,s^{-2}}) has value 6 N6\,\mathrm N: a force unit, not an acceleration unit.

Watch out

Writing “weight =3 kg=3\,\mathrm{kg}” uses a mass unit for a force. State mass in kilograms and weight in newtons; a correct numerical value without the right quantity is not enough.

Try a fresh set of readings over a longer time interval.

Convert before substituting

The controller in the opening question uses centimetres and minutes. Each of those unit choices must be converted, including every power.

For comparison, a speed of 72 km h−172\,\mathrm{km\,h^{-1}} becomes 72×1000/3600=20 m s−172\times1000/3600=20\,\mathrm{m\,s^{-1}}. The factor 10001000 changes kilometres to metres and the factor 36003600 changes hours to seconds.

Unit factors and powers

1 cm min−1=0.0160 m s−11 cm min−2=0.01602 m s−2\begin{aligned} 1\,\mathrm{cm\,min^{-1}}&=\frac{0.01}{60}\,\mathrm{m\,s^{-1}}\\ 1\,\mathrm{cm\,min^{-2}}&=\frac{0.01}{60^2}\,\mathrm{m\,s^{-2}} \end{aligned}

Convert a powered unit by applying the same power to its conversion factor. The factors here are exact.

Not in the formula sheet: learn it.

The difference between a speed and an acceleration can turn one factor of 6060 into 36003600. That is why the first controller's enormous number is smaller than it looks.

Continue this lesson.

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