Lesson Video: Quantities and Units in Mechanics Mathematics

Video Transcript

In this video, we will learn how to identify fundamental and derived quantities used in mechanics, such as length, time, and velocity, and identify their units and unit conversions.

A unit is a defined magnitude of a quantity. Any other magnitude of the same quantity may be expressed as a multiple of the unit. For example, the kilogram is a unit of mass. And any other mass, such as the mass of a typical person, may be defined as a multiple of that unit. Specific units of measurement are arbitrary and may be defined by convenience. For example, degrees and radians were arbitrarily defined to be useful in certain situations.

Radians are typically more useful in mathematical problems, such as trigonometry, since the base functions of sine and cosine operate naturally with radians. Quantities in degrees, minutes, and seconds must first be converted into radians before inputting them into trigonometric functions. But they are a much more intuitive way for humans to think about angles. Nevertheless, both of these units describe the same quantity, also known as a dimension, in this case an angle.

Every system of units has a unit for every physical quantity. And the most commonly used system of units is the International System of Units, or SI. The International System of Units consists of seven base units for the following base quantities: for mass, the kilogram or kg; for time, the second, s; for length, the meter, m; for electric current, the ampere, capital A; for amount of substance, the mole, mol; for luminous intensity, the candela, cd; and for thermodynamic temperature, the Kelvin, capital K. Each of these base quantities also comes with a dimension symbol, which is used in dimensional analysis, which we’ll get to in a moment.

The International System of Units forms a coherent system of quantities, meaning that any other physical quantity can be expressed as a product of powers of any of the seven base units. So let’s look at an example for how to derive the SI unit for a quantity that is not among the seven base units.

Velocity is equal to the distance covered over a period of time. Which of the following is not a unit of velocity? Is it (a) centimeters per second, (b) meters per second, (c) meters per second squared, (d) meters per minute, or (e) kilometers per hour?

Since velocity is equal to the distance covered over a period of time, velocity 𝑣 is given by the equation 𝑣 equals 𝑑 over 𝑡, where 𝑑 is the distance and 𝑡 is the time elapsed. We can determine the dimension of velocity by performing dimensional analysis on this equation. All this means is replacing every variable on the right-hand side of the equation with its dimension. So, in this case, the distance 𝑑 has dimension length 𝐿 and the time elapsed 𝑡 of course has dimension time 𝑇. Both sides of the equation must have the same dimension. Therefore, the dimension of 𝑣 must be the same as the dimension of 𝑑 over 𝑡, 𝐿 over 𝑇.

We can now compare this with every one of our answers to see which one does not have units representing dimension L over 𝑇. So, for answer (a), we have the centimeter, which is a multiple of the meter, the base unit for length. And we have second, which is the base unit for time. So answer (a) is 𝐿 over 𝑇. So this is indeed a unit for velocity.

For answer (b), we just have the base unit for length over the unit for time. So this is also a unit for velocity. For answer (c), we have the base unit for length, the meter, divided by the base unit for second squared. So the dimension is 𝐿 over 𝑇 squared. Therefore, (c) is not a unit for velocity. For completeness, for answer (d), we have the base unit for length, again, divided by the minute, which is a multiple of the second. So we have length over time once again. And likewise, for (e), we have the kilometer, a multiple of the meter, divided by the hour, a multiple of the second. So we once again have 𝐿 over 𝑇. Therefore, our answer is (c). Meters per second squared is not a unit of velocity.

Sometimes there may be more than one step to deriving the SI unit for a quantity not among the seven base quantities. Such quantities and units are often given a special name. The International System of Units contains a further 22 derived quantities. Here are a few examples. As you can see, many of these are very fundamental quantities with units you may already be familiar with, such as the newton for force. Some of them are even surprising, such as the coulomb for electric charge. One might think that electric charge is a more fundamental quantity than current. But it turns out that the electric current is easier to quantify in practice.

All of these derived units may be written as a product of powers of the SI base units. So, for example, the newton may be defined as one kilogram raised to the power of one times one meter raised to the power of one divided by one second squared. Some quantities, such as plane angle, are in fact what we call dimensionless or dimension one. This is because they are products of the SI base units all raised to the power of zero.

Taking the plane angle as an example, we can think of an angle as a kind of proportion. For example, the angle 𝜋 radians is equal to one-half of the circle, or one-half of the circle’s circumference, divided by its full circumference. Both of these quantities are lengths. So the radian, in fact, has SI base units of meters over meters or meters to the zero. And of course anything raised to the power zero is just one. Some of these units, for example, the volt for electric potential difference, can get fairly complicated in terms of the SI base units. Let’s look at an example of how to derive these more complex units from the seven base units.

The kinetic energy, which is measured in joules, is given by the rule 𝑇 equals one-half 𝑚𝑣 squared. Which of the following units is equal to the joule? (a) Kilogram per meter per second squared. (b) Kilogram times meter per second. (c) Kilogram times meter squared per second squared. (d) Kilogram per meter squared per second squared. Or (e) kilogram times meter per second squared.

Consider the equation for the kinetic energy: 𝑇 equals one-half 𝑚𝑣 squared, where 𝑇 is the kinetic energy, 𝑚 is the mass, and 𝑣 is the velocity. In an equation representing physical quantities, the dimensions of the quantities on both sides must be the same. Therefore, the dimension of the kinetic energy 𝑇 must be equal to the dimension of one-half times the dimension of 𝑚 times the dimension of 𝑣 squared. One-half is just a number. Therefore, it is dimensionless, or dimension one. 𝑚 is just mass, so it has dimension mass. Velocity squared is a little more complicated since it is not a base SI dimension.

To find velocity squared’s dimension in terms of the SI base dimensions, we will need to consider the equation for velocity as well. Recall that the velocity is given by the displacement 𝑑 divided by the time elapsed 𝑡. We now have velocity expressed purely in terms of base SI quantities, length and time. The displacement has dimension length 𝐿 and the time elapsed has dimension time 𝑇, not to be confused with the 𝑇 for kinetic energy. Therefore, 𝑣 squared equal to 𝑑 squared over 𝑡 squared has dimension 𝐿 squared over 𝑇 squared. Putting all these together in the original equation for kinetic energy, we get the dimension of kinetic energy 𝑇 is equal to one times 𝑀 times 𝐿 squared over 𝑇 squared.

We can now substitute in the SI base unit for each of these quantities. Dimension one has no specific unit. For mass 𝑀, we have the kilogram. For length 𝐿, we have the meter, in this case squared. And for time 𝑇, we have the second, in this case also squared. Putting all these together, we get the unit for kinetic energy in terms of the SI base units: kilogram meter squared per second squared. Comparing this with our possible answers, we can see this matches with (c) kilogram times meter squared per second squared.

Sometimes it might not be convenient or helpful to express the unit of a quantity in terms of the SI base units. For example, the torque 𝜏 represents the turning effect of a force and is given by the equation 𝜏 equals 𝐅𝑑, where 𝐅 is the force and 𝑑 is the perpendicular distance from the pivot point. If we were to perform dimensional analysis on this equation to find the SI unit for torque, we would start by noting that 𝐅 has dimension force and 𝑑 has dimension length, the latter being an SI base dimension. To find the dimension for force, we would start with Newton’s second law, force equals mass times acceleration, then convert this to mass times velocity over time, and convert this to mass times distance over time over time.

Simplifying the right-hand side, we would get 𝑚𝑑 over 𝑡 squared, which is entirely in SI base quantities and has dimension 𝑀𝐿 over 𝑇 squared. Therefore, the dimension of force in terms of SI base dimensions is 𝑀𝐿 over 𝑇 squared. So force has an SI base unit of kilogram meter per second squared.

Substituting the dimension of force into the equation for torque, we get the dimension for torque equal to 𝑀𝐿 squared over 𝑇 squared. Therefore, the unit for torque in terms of the SI base units is kilogram meter squared per second squared. This is exactly the same unit as the unit for energy, even though the two quantities are entirely different. This shouldn’t be too surprising though, since in a different context force times distance also gives the work done, or energy. This is partly why we choose to give derived quantities, such as energy, special names for their units, since it contextualizes the unit.

Torque does not in fact have its own special unit. And the most common convention is to use the newton-meter, one newton times one meter. Even though these are technically the same units, whenever we see joules, we are usually talking about energy or work done. And whenever we see newton-meters, we are usually talking about moments or torque.

Another common convention is to use units with different magnitudes from the SI base units. For example, the SI base unit for speed is meters per second. But in different contexts, for example, the speed of a car, a different unit might be used for convenience, like kilometers per hour. However, it’s often necessary to work with quantities in different units. In such cases, we usually convert all quantities into SI base units first. Let’s look at an example of how to convert units of different magnitudes into SI base units.

What is 2.87 times 10 to the seven dynes in newtons?

The dyne is defined as one gram times centimeter per second squared and is particularly useful when dealing with very small forces, such as surface tension. The newton is a derived SI unit and is given in terms of the SI base units as one kilogram times meter per second squared. We can now proceed by converting all of the units on the right-hand side of the equation for the dyne into SI base units.

So to start with, one gram is one kilogram divided by 1000, which is equal to 10 to the negative three kilograms. And similarly, one centimeter is equal to one meter divided by 100, which is equal to 10 to the negative two meters. We already have the last term, second squared, in terms of the base SI units. So we don’t need to do anything. Substituting these conversions into the equation for the dyne, we get one dyne equals one times 10 to the negative three kilograms times 10 to the negative two meters over second squared. Collecting powers of 10, we get 10 to the negative five kilogram meter squared per second squared. So we have one dyne equals 10 to the negative five newtons.

If we now multiply both sides of the equation by 2.87 times 10 to the seven, we get the number of dynes in terms of newtons. Simplifying the right-hand side, we get our final answer. 2.87 times 10 to the seven dynes equals 287 newtons.

Sometimes converting into SI base units may not be as straightforward as multiplying by a power of 10. In the final example, let’s look at how to convert from units with nonstandard magnitudes to SI base units and vice versa.

Which of the following is false? (a) 72 kilometers per hour equals 20 meters per second. (b) Three kilometers per minute equals 50 centimeters per second. (c) 15 meters per second equals 54 kilometers per hour. Or (d) 42 meters per minute equals 70 centimeters per second.

To verify each equation, we need to convert the units on one side of the equation into the units on the other side of the equation. Let’s convert the units from the left-hand side of each equation into the units on the right-hand side of each equation and then compare the coefficients to see if they are true or false.

So, starting with (a), we need to convert kilometers into meters and hours into seconds. One kilometer is equal to 1000 meters, and one hour is equal to 3600 seconds. We therefore have 72 kilometers per hour equals 72 times 1000 meters per 3600 seconds. It can be helpful to write these as fractions so we can see the multiplication and division more clearly. So we have 72000 meters divided by 3600 seconds. Simplifying this, we obtain 20 meters divided by one second, or 20 meters per second. Therefore, equation (a) is true.

Next, for equation (b), we need to convert kilometers into centimeters and minutes into seconds. One kilometer equals 1000 meters. So in centimeters, one kilometer is 100000 centimeters, or 10 to the five centimeters. And similarly, one minute equals 60 seconds. Therefore, we have three kilometers per minute equals three times 10 to the five centimeters per 60 seconds. Simplifying this fraction, we get three kilometers per minute equals 5000 centimeters per second. Therefore, the false equation is (b) three kilometers per minute equals 50 centimeters per second.

Let’s finish this video by recapping some key points. The International System of Units, or SI, consists of seven base units for seven base physical quantities. Every other physical quantity has a unit that may be expressed as a product of powers of these seven base units. The unit for any physical quantity in terms of the SI base units may be found from an equation for that quantity using dimensional analysis. And finally, units of nonstandard magnitudes may be converted to SI units by multiplying by the appropriate factor. For example, one hour equals 3600 times one second.