3.01a–3.01c - Mechanical quantities, SI units and dimensions

3.01a–3.01c - Mechanical quantities, SI units and dimensions

Turn a measurement into something you can use: choose the independent SI quantities in 3.01a, build and convert the derived units in 3.01b, and use the moment unit in 3.01c. You will also learn how units can expose a wrong equation before you calculate.

Independent base quantities

A robot travels at 5400 cm min−15400\,\mathrm{cm\,min^{-1}}. Another travels at 0.8 m s−10.8\,\mathrm{m\,s^{-1}}. Which is faster? The larger number is not enough to decide: first we need a common measuring language.

A robot travelling 2 m2\,\mathrm m has travelled 200 cm200\,\mathrm{cm}. The measurement has changed its numerical form, while the length itself has stayed the same.

You already know

A measurement combines a number and a unit. Multiplying by a conversion factor equal to one changes the unit without changing the quantity.

Mechanics starts with three different kinds of measurement. Knowing one does not supply either of the others.

Independent base quantities

The SI base quantities used here are length, time and mass, measured in metres (m)(\mathrm m), seconds (s)(\mathrm s) and kilograms (kg)(\mathrm{kg}). They are mutually independent: none can be expressed using just the other two.

Two trolleys can have the same length and mass but take different times to cross a room. An equation can relate these quantities when extra physical information is supplied; it cannot convert kilograms into seconds.

Mass describes the amount of matter in a body and remains a mass even when the body is moving. A 750 g750\,\mathrm g parcel has mass 0.750 kg0.750\,\mathrm{kg}. Its weight is a force, measured in newtons, so a mass and a weight are different quantities.

Key idea

A quantity is what you measure; a unit is the agreed size used to report it. Changing the unit changes the number, not the physical quantity.

Use that distinction to test a claim about the robot.

Units built from relationships

A derived unit records the operations connecting quantities. Read those operations and the unit becomes useful information, rather than a label to memorise.

If displacement changes by 12 m12\,\mathrm m in 4 s4\,\mathrm s, its average rate of change is 3 m s−13\,\mathrm{m\,s^{-1}}. Velocity includes direction; speed is its non-negative magnitude. They have the same unit, just as two different kinds of length can both be measured in metres.

Derived mechanical 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 on a body.

For motion along a line, choose a positive direction. A negative velocity describes motion in the opposite direction; it is not a negative speed. The force–acceleration relationship identifies the newton's base units here; solving motion problems with that relationship comes later.

SI derived units

velocity unit=m s−1acceleration unit=m s−21 N=1 kg m s−2\begin{aligned} \text{velocity unit}&=\mathrm{m\,s^{-1}}\\ \text{acceleration unit}&=\mathrm{m\,s^{-2}}\\ 1\,\mathrm N&=1\,\mathrm{kg\,m\,s^{-2}} \end{aligned}

Force and weight are measured in newtons. These units follow from displacement divided by time, velocity divided by time, and mass multiplied by acceleration respectively.

Not in the formula sheet: learn it.

Start with a velocity change whose direction is already described by signs.

Worked example 1: attach the acceleration unit

Question. A trolley's velocity changes from 5 m s−15\,\mathrm{m\,s^{-1}} to 14 m s−114\,\mathrm{m\,s^{-1}} in 3 s3\,\mathrm s. Find its average acceleration, including its unit.

Plan. Divide the change of velocity by the elapsed time, applying the same division to the units.

Δv=14−5=9 m s−1final minus initialaverage acceleration=93=3 m s−2divide by seconds\begin{aligned} \Delta v&=14-5=9\,\mathrm{m\,s^{-1}} &&\text{final minus initial}\\ \text{average acceleration}&=\frac{9}{3}=3\,\mathrm{m\,s^{-2}} &&\text{divide by seconds} \end{aligned}

Why this step? Dividing a velocity by time introduces a second factor of s−1\mathrm{s^{-1}}.

Answer. The average acceleration is 3 m s−23\,\mathrm{m\,s^{-2}} in the positive direction.

Check. A change of 3 m s−13\,\mathrm{m\,s^{-1}} each second for 3 s3\,\mathrm s gives the stated total change of 9 m s−19\,\mathrm{m\,s^{-1}}.

Now change only the order of the endpoint velocities. The unit stays the same; the sign should change.

Worked example 2: retain the sign

Question. A trolley's velocity changes from 14 m s−114\,\mathrm{m\,s^{-1}} to 5 m s−15\,\mathrm{m\,s^{-1}} in 3 s3\,\mathrm s. Find its average acceleration, including its unit.

Plan. Use final minus initial velocity, then divide by the elapsed time.

Δv=5−14=−9 m s−1same positive directionaverage acceleration=−93=−3 m s−2divide by time\begin{aligned} \Delta v&=5-14=-9\,\mathrm{m\,s^{-1}} &&\text{same positive direction}\\ \text{average acceleration}&=\frac{-9}{3}=-3\,\mathrm{m\,s^{-2}} &&\text{divide by time} \end{aligned}

Why this step? A direction convention controls the sign; the word “acceleration” does not require a positive answer.

Answer. The average acceleration is −3 m s−2-3\,\mathrm{m\,s^{-2}}.

Check. The final velocity is smaller by 9 m s−19\,\mathrm{m\,s^{-1}}, so the negative sign fits.

The same unit discipline distinguishes a force from a mass. A product (2 kg)(3 m s−2)(2\,\mathrm{kg})(3\,\mathrm{m\,s^{-2}}) is 6 N6\,\mathrm N, not 6 kg6\,\mathrm{kg}.

Watch out

The line “acceleration =3 m s−1=3\,\mathrm{m\,s^{-1}}” gives a velocity unit. Acceleration measures velocity change per second, so write 3 m s−23\,\mathrm{m\,s^{-2}}.

Try the same method with a scooter and a longer time interval.

Convert before substituting

A fast way to make a slow robot look impressive is to quote its speed in centimetres per minute. Converting both the numerator and denominator brings the comparison back to earth.

Since 1 km=1000 m1\,\mathrm{km}=1000\,\mathrm m and 1 h=3600 s1\,\mathrm h=3600\,\mathrm s, the speed 54 km h−154\,\mathrm{km\,h^{-1}} is 54×1000/3600=15 m s−154\times1000/3600=15\,\mathrm{m\,s^{-1}}. A numerical speed in kilometres per hour is divided by 3.63.6 to express it in metres per second.

Convert the whole unit

1 cm min−1=0.0160 m s−1,1 cm min−2=0.01602 m s−21\,\mathrm{cm\,min^{-1}}=\frac{0.01}{60}\,\mathrm{m\,s^{-1}}, \qquad 1\,\mathrm{cm\,min^{-2}}=\frac{0.01}{60^2}\,\mathrm{m\,s^{-2}}

A power on a unit applies to its conversion factor too. These are exact conversion factors.

Not in the formula sheet: learn it.

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