A review of centimeter-gram-second and foot-pound-second units, with some dynamics thrown in

- Introduction
- The cgs System
- Gravity, Density, Force and Mass
- The fps System
- Integrals of the Equations of Motion and Conservation Laws
- Dynamics in Polar Coordinates
- Inertial Systems, Rotating Systems and the Coriolis Force
- Friction and Viscosity
- Illustrative Examples
- Physical Constants in CGS
- References

Scientific and engineering work these days is mostly done in *Système Internationale* (SI) units, otherwise known as the MKS system, based on the metre, kilogram and second as the fundamental units, and enforced by the Unit Police that tell us what to do through many regulations and deprecations. The reader may encounter the good old units based on the centimetre, gram and second (cgs), or on the foot, pound and second (fps), and wonder what they are about. The cgs system was once standard in Physics, the fps system in Engineering in Britain and the United States, and to use them strikes a blow for freedom and tradition. They are, of course, just as good as SI, and not enforced by tedious pedants. This article reviews the two systems, illustrated by a review of the Newtonian mechanics of particles.

An *absolute* system of units lets the units or "dimensions" of all quantities be determined by the defining equations, without any arbitrary choices. Arbitrary choices were once the fashion, in order to have convenient units for particular activities. For example, we have leagues, miles, chains, rods, perches, fathoms, yards, feet, inches and barleycorns, all expressing distance in English traditional units. A similar variety is available for most other quantities in daily use. The benefits of having units suited to the task at hand should not be depreciated, but when we are doing physics it is extremely tedious to put in conversion factors for each equation, and often obscures what we are studying. For this purpose, we require an absolute system of units. After our calculations are done, we can convert the results to any convenient unit. The millimeter is an excellent unit for things like optics and machining, but to say "give him and inch, and he'll take 1,609,300 millimeters (approximately)" does not appeal. "How far is it to Boulder?" you may be asked. If you answer 5.15 x 10^{7} mm, your friend will still be effectively in the dark if she is wondering how long it will take to drive there. If you ask her how old she is, an answer of 9.5 x 10^{8} seconds will not give an immediate impression. We use kilometers and years for good reasons. Granted, SI does use decimal multiples that are often not inconvenient (like the millimeter), and conversions depending only on moving the decimal point are especially convenient. We now have calculators, so this benefit is somewhat blunted.

The SI is more French than it is scientific. The Celsius temperature scale is included, but it is no more metric or decimal than the good old English Fahrenheit scale or the old German Réaumur scale. There are exactly 100° between 0° and 100° in each system, the cynical may note. All the scales are arbitrary. It is easier to remember the freezing and boiling temperatures of water at a standard atmospheric pressure in Celsius, of course, but this is not of pressing significance to most people. The metre was designed when the French still thought the earth was perfectly spherical, and made it 1/10000 of a quadrant. This was a perfectly silly definition, impossible to measure accurately. When sphericity of the earth proved erroneous, the metre became the length of a bar in Paris, just another arbitrary unit. The seconds pendulum is about 1 m long, and this was probably what really suggested the metre. The definition of the kilogramme as the mass of a 10-cm cube of water was another poor definition, impossible to measure accurately. Water and cubes expand and contract thermally. The kilogramme became the mass of a lump of metal in Paris. The second was still 1/86400 of a mean solar day, somewhat better but still not easy to determine with precision. The French originally wanted it to be something like 1/100000 of a mean solar day, but redefining people's time (with 10 months in a year, and other absurdities) was not on. So metric units, which became the SI, were just another arbitrary choice. The French public, incidentally, had to be forced by law to use the metric system when they did not embrace it, sticking to *toise*, *pouce* and *livre* until threated by violence. Things have not changed a lot. Note that I use the French spelling metre, kilogramme, since this is required by the SI standards. Nobody outside France seems to use "seconde" though. Enough is enough.

The cgs system is closely related to SI, differing only by powers of 10, since the standard units are centimeter (L), gram (M) and second (T), where we represent the "dimensions" by the letters in parentheses. The SI unit of force is the newton (N), and its dimensions are MLT^{-2}. To find the equivalent in cgs, multiply by 1 = (1000g/1kg)(100cm/1m) and find 1 N = 10^{5} cgs units of force. The cgs unit of force is the *dyne* (from the Greek for "force"), so we have 1 N = 10^{5} dyne. Sometimes "dy" is used as an abbreviation for "dyne." The reader is probably quite familiar with this easy way of converting with unit ratios, which always works. The SI unit of work or energy is the joule (J), which is defined as N-m (W = fd). Its dimensions are ML^{2}T^{-2}. Then 1 J = 10^{5} dyne x 100 cm = 1 x 10^{7} erg, where "erg" (from the Greek for work) is the cgs unit's name. Pressure is force per unit area, dimensions M^{-2}LT^{-2}. The SI unit is the pascal (Pa), 1 N/m^{2}. Then, 1 Pa = 10^{5}(100)^{-2} dy/cm^{2} = 1000 dy/cm^{2}. The cgs unit of pressure does not have a special name. It happens that the normal atmospheric pressure at sea level is about 10^{7} Pa, which is called a *bar*. That pressure is really about 1.01325 bar. The *millibar*, often seen in weather reports, is then 10000 dy/cm^{2}. In traditionally metric countries, the kg/cm^{2} is a common unit of pressure, which is also about 1 atm.

Newton's Second Law states that the acceleration of a body is equal to the ratio of the force acting on the body to its mass, f/m = dv/dt. This elegant statement introduces two new quantities, force and mass, neither of which is independently defined, and both of which have been the subjects of extended philosophical discussion. If mass is a fundamental unit, then [f] = MLT^{-2} (the square brackets signify "dimensions of"). If force is taken as a fundamental unit, then [m] = FL^{-1}T^{2}. Mass seems to be a fundamental quality of a body, while force appears to be accidental, so mass is a reasonable choice as a fundamental unit. On the other hand, mass cannot be experienced directly, while force is constantly encountered in practice, so a case can also be made for force as a fundamental unit.

Newton's Law of Gravitation does give a certain force directly, the gravitational force f = Gmm'/r^{2} between two masses. If m and m' are in gram, r in cm, then G = 6.672 x 10^{-8} dy-cm^{2}/g^{2}. The universal gravitation constant G is not known to great precision because it is equivalent to weighing the earth--not a convenient thing. It, too, gives only a ratio of force to mass, since f/m = Gm'/r^{2}. If r is the radius of the earth, 6378 km, and m' its mass, 5.98 x 10^{27} g, then f/m at its surface is g = 981 cm/s^{2}, called the acceleration due to gravity. This is a convenient standard acceleration for us to use to relate force and mass, where now the force is called *weight*, to imply the standard acceleration. Actually, we measure g and use it to determine the mass of the earth, of course.

A problem with this is that g varies with position. At any one point, it is constant enough, but has different values at different points, which gives the same mass different weights at different places. The g we observe is the joint result of the gravitational attraction of the earth and the centrifugal acceleration due to its rotation. At the equator, g is about 978.039 gal ("gal" stands for Galileo, and is used in gravitational studies for cm/s^{2}, together with the smaller milligal = 10^{-3} gal). Here we are furthest from the center of the earth, and the centrifugal force is greatest. At the pole, g = 983.217 gal. The difference is about 0.5%. We are closest to the center of the earth, and the centrifugal force is zero. At any latitude φ, g = 978.049(1 + 0.0052884 sin^{2} φ + 5.9 x 10^{-6} sin^{2} 2φ) gal. Here, Heiskanen's 1928 value for the equatorial gravity is used. If we are above sea level, the *free-air* correction 0.3086h mgal must be subtracted, where h is the altitude in metres. There are other corrections depending on the masses above and around the points, and when all this has been considered, there are still "anomalies" of up to about 100 mgal that cannot be explained by anything obvious. All this is to show that the actual acceleration of gravity is not a uniform value. The value g = 981 cm/s^{2} is a good value for everyday use.

Since we know the mass and the radius of the earth, we may itch to know its average *density*, or ratio of mass to volume. Its density d = m/(4πr^{2}/3) = 5.5 g/cm^{3}. The ratio of the mass of a body to the mass of an equal volume of water is called its *specific gravity*, sp.gr., and in the cgs system is numerically equal to the density, but is dimensionless (and so the same in any system of units). The specific gravity of granite is 2.64-2.76, of dolomite 2.84, of limestone 2.68-2.76, of sandstone 2.14-2.36, of feldspar 2.55-2.75 and of mica 2.6-3.2. A good characteristic value for surface rocks is about 2.7. The earth can't be made up of surface rocks! Hematite, an iron ore, has a sp.gr. of 4.9-5.3. Rock salt, halite, is 2.18, glass 2.4-2.8, bituminous coal 1.2-1.5, crude oil 0.8, oak 0.7 and pine 0.5. The sp.gr. of aluminum is 2.8, zinc 7.1, cast iron 7.2, steel 7.9, brass 8.5, copper 8.9 and lead 11. The sp.gr. of ice is 0.917, so it floats. The sp.gr. of water is 1.00 at 3.98°C, 0.99823 at 20°C, and 0.95838 at 100°C. Mercury, used in barometers and such, has 13.5953 at 0°C and 13.5462 at 20°C. Dry air has a density of 1.2929 g/liter at 0°C, and its density at any other temperature and pressure can be estimated from the gas laws. These are representative values that show the range for common materials, and can be used in approximate calculations.

If we must relate mass and force, however, some such value must be assumed as a constant conversion factor. It is no longer the actual acceleration of gravity, but an arbitrary constant. Nevertheless, it should be close to an average value of g. The value 980.665 gal has been chosen to fill this role. It would have been just as easy to have chosen 981 gal, but no, we have to make it look like something precise. I think this is the value measured at Potsdam some years ago. When we use this value, try to eliminate any suggestion in your mind that it involves some kind of physical measurement. It is just an arbitrary conversion constant.

The absolute foot-pound-second system of units (fps) retains the second as the unit of time. The unit of length is the foot, 30.48 cm (exactly). Whoever decided that an inch should be exactly 2.54 cm has my thanks. Another definition, 1 metre = 39.37 in, which gives 1 inch = 2.54000508 cm, was used to define the *US survey foot*, used in geodesy. There is no difference practically, only an annoyance in measurements of high precision. The third fundamental unit is the pound, the unit of *force*. The Handbook of Chemistry and Physics tells me that 1 pound = 0.45359237 kg. Now kg is mass, so this really means that 1 pound force = 1 lbf = (0.45359237)(980.665) dyne = 4.4482216 x 10^{5} dynes = 4.4482216 N, or, that a pound force is the weight of 453.59237 g. Someone went to a lot of trouble to compare the standard kilogram in Paris with the standard pound in London, and came up with all those digits for us to admire. A choice of 1 lb = 454 g would have done just as well to define the fps system, with only a negligible amount of restandardization. The standard acceleration of gravity in the fps system is 980.665/30.48 = 32.17405 ft/s^{2}. In everyday problems, the value 32.2 ft/s^{2} is normally used. The foot-pound is 1.35582 J.

It is usual to refer to the weight of a body in the fps system, not its mass. One cc of water has a mass of 1 g, and so a weight of 1/453.59237 lb in fps. A cubic foot of water is 28316.8 cc, so it weighs 62.428 lb at 3.98°C or 62.317 lb at 20°C. The figure of 62.4 is commonly accepted by engineers. Using the specific gravity, we can now find the weight of a cubic foot of anything. Since w/m = g, m = w/g in the fps system, so we have f = (w/g)dv/dt for Newton's Law. The appearance of g caused the name "gravitational system" to be applied to this kind of fps system, but it is a misnomer. There is nothing gravitational about it, and it does not depend in any way on the acceleration of gravity. The "g" is only a conversion constant that happens to be approximately equal to the acceleration of gravity. It is *never* anything other than 980.665. The unit of mass has been named the *slug*. The weight of a slug is about 32.17 lb. It is equal to 14.593 kg. Mass densities should be specified as slug/ft^{3}. Civil engineers adopted the slug, mechanical engineers didn't appreciate the problem, and chemical engineers did something else which I shall mention in a moment. The symbol γ is used for weight density, γ = w/V in, for example, pounds per cubic foot, pcf. Mass density is still ρ or d.

There are two general ways to determine the weight or mass of a body. One way is to use a *balance*, where the mass is compared with a known mass using levers or the equivalent. This includes the once-common laboratory beam balance. The result is unaffected by variations in gravity. The other way is with a device that measures the force directly, such as a spring scale or strain gauge. Such a *spring scale* must be calibrated, and the calibration will depend on the local value of gravity. Household spring scales are not sensitive enough to detect variations in gravity. Most small "balances" are now electronic spring scales.

There are other ways of defining fps systems. One way lets the unit of mass be the pound, or lbm. Then f = mdv/dt defines the unit of force, called a *poundal*, pdl. The force acting on 1 lbm in standard gravity is, then, 32.2 pdl, which must be its weight. 1 pdl = 13,810 dynes, we can easily discover. This system was used in school physics for many years, but never really caught on except with physics teachers. Another way keeps both the pound mass (lbm) and the pound force (lbf), which means that Newton's law has to be written f = (m/g)dv/dt, causing all kinds of mischief. Note that this is not just f = (w/g)dv/dt, which is harmless, since w/g is just mass. By contrast, m/g isn't anything holy. Here, g is not just a conversion factor, but is an essential part of the equation, contributing units as well as magnitude. Chemical engineers used this system. Since they don't use Newton's Second Law much, it doesn't do a lot of harm, and allows the carefree confusion of two different units with the same name.

In using the fps system, it is best to use feet and pounds consistently, and to divide them decimally, just like metres and kilograms. If it is not otherwise obvious, whether factors of g should be introduced can be determined from the units. For convenience, things like inches, miles, tons and acres can be brought in later, but not used in theoretical calculations. For example, standard atmospheric pressure is 2117 psf, and water weighs 62.3 pcf, air 0.075 pcf. A speed of 30 mph is 44 fps. Quantities can be expressed in consistent units by factors such as these before performing a calculation.

It is a subsidiary matter that the cgs system generally uses the Celsius thermometer, and the Kelvin for absolute temperature, K = °C + 273.15. The fps system generally uses the Fahrenheit thermometer and degrees Rankine for absolute temperature, °R = °F + 491.69. Of course, 5K = 9°R. Either temperature scale can be used with either system of units. Specific heats are numerically the same in cal/K and Btu/°R. 778 ft-lb = 1 Btu, and 4.186 J = 1 cal. The gas constant is 8.31441 x 10^{7} erg/K-gmol, or 1545.33 ft-lb/°R-lbmol.

Let's now look at integrals of the equations of motion. If we multiply by dt and integrate, we find ∫(1,2)fdt = ∫m(dv/dt)dt = m∫(1,2)dv = mv_{2} - mv_{2} = Δmv. ∫fdt is called *impulse*, and mv is the *momentum*. Neither of these quantities has any special name, but the units are ML/T: gm-cm/s, dyne-s or lb-s. Multiplying by dx instead, ∫(1,2)fdx = ∫(dv/dt)dx = ∫(dv/dx)(dx/dt) = ∫vdv = mv_{2}^{2} - mv_{1}^{2} = Δ(mv^{2}/2). These quantities are much used: ∫fdx is *work*, and mv^{2}/2 is *kinetic energy* T. In cgs, the unit of mechanical energy is the erg (from the Greek for work), the dyne-cm, and in fps, it is the ft-lb. The MKS unit is joule = N-m, so 1 J = 10^{5} x 10^{2} = 10^{7} erg. The time rate of change of kinetic energy is dT/dt = mv(dv/dt) = fv, called *power*. It can be expressed as erg/s or ft-lb/s, and 10^{7} erg/s = 1 W. An erg/s is 0.1 μW. In fps, we may use the colorful horsepower, 1 hp = 550 ft-lb/s. Since 1 ft-lb = 1.356 x 10^{7} erg, 1 hp = 745.7 W.

If f is a function of position, f = f(x,y,z), and ∫fdx around any closed curve is zero (as it is in the gravitational field of the earth), then it a function V(x,y,z) can be found such that f_{x} = -∂V/∂x, or, in vectors, **f** = -grad V. This procedure is very familiar from electrostatics. The function V is the *potential energy*. Then, ∫(1,2)fdx = -∫(dV/dx)dx = V_{1} - V_{2} = -ΔV. Combining this with our earlier expression, we have ΔV + ΔT = Δ(T + V) = 0, or T + V = E = constant. This is the principle of *conservation of mechanical energy*, which proves very useful.

Energy leads the way into advanced dynamics, where it replaces the concept of force. For example, the potential energy function V = kx^{2}/2 yields the force f = -kx, which we know gives us the harmonic oscillator. We use Newton's Law to give -kx = md^{2}x/dt^{2}, or x" + (k/m)x = 0, where the prime indicates differentiation with respect to t. The solution of this equation is x = A sin ωt + B cos ωt, with A and B constants that are determined by the initial conditions, and ω = √(k/m). In advanced dynamics, we form the Lagrangian L = T - V and use Lagrange's equations to get the equations of motion. This is only added complication in this simple case, but the method is capable of great generalization.

Another important conservation law comes from Newton's Third Law, which states that if one body exerts a force on another, then the second body exerts an equal and opposite force on the other. This says that if **f** = d**p**_{2}/dt, then -**f** = d**p**_{1} /dt, from which it follows that d(**p**_{1} + **p**_{2})/dt = 0, so that **p**_{1} + **p**_{2} = constant. This is the principle of the *conservation of momentum*. Each component of the momentum is separately conserved. The conservation of energy and momentum are powerful tools for the study of collisions. An *elastic* collision is one in which the energy is conserved. Even in an inelastic collision, the momentum is still conserved.

If the force is constant, then we can integrate the equation of motion twice and find the position as a function of time. With (f/m) = a, we find x = at^{2}/2 + v_{0}t + x_{0} with two constants, the initial velocity and position. This solution is studied in every elementary physics course, and is applicable to the everyday problem of projectile motion under gravity. There are many useful consequences of this solution, such as v^{2} - v_{0}^{2} = 2a(x - x_{0}), where the time has been eliminated between it and its first derivative. In the form v = √2as it is very useful, even in things like hydrodynamics.

So far we have not emphasized the vector nature of Newton's Second Law, since in rectangular coordinates we can write one equation for each dimension and work entirely with components. In fact, many practical problems are one-dimensional, and the subject does not come up. When we consider motion in a plane described by polar coordinates, however, we must look more deeply. If a point is given by P(r,θ), then the unit vector **r'** is **r**/r, and the unit vector **θ'** is at right angles to it toward the direction of increasing θ, as shown in the diagram at the right. The primes will have to do here for identifying the unit vectors. Now, since **r** = r**r'**, **v** = d**r**/dt = (dr/dt)**r'** + r(d**r'**/dt) = (dr/dt)**r'** + rω**θ'**, since d**r'**/dt = ω**θ'**, where ω is dθ/dt. Note also that d**θ'** = -ω**r'**, because we need this for the acceleration: **a** = [d^{2}r/dt^{2} - rω^{2}] **r'** + [2(dr/dt)ω + r(d^{2}θ/dt^{2})]**θ'**. This expression has many applications.

Suppose first that the force acting on a particle moving in a plane is *central*, that is, directed toward the origin. Then the θ component of the acceleration must vanish, or 2(dr/dt)ω + r(dω/dt) = 0. If this is multiplied by r, we find that d(r^{2}ω)/dt = 0, or r^{2}ω = h, a constant. Then, mr^{2}ω = mv_{p}r = H, a constant. Here, v_{p} is the component of velocity perpendicular to the radius. In vector notation, **H** = m**r** x **v** is called the *angular momentum*, the moment of the linear momentum about the origin. With a central force, then, the angular momentum is a constant. One consequence of this is Kepler's Area Law.

Now suppose the particle is moving in a plane with constant r, as shown in the figure. This is the case for all the points in a rigid body rotating about fixed axis normal to the plane. Then **a** = -rω^{2}**r'** + rdω/dt**θ'**. The first term is an acceleration directed toward the origin of amount v^{2}/r, the *centripetal acceleration* that must be supported by a force directed toward the orgin. The second term is zero unless there is an *angular acceleration*, which requires a force in the direction of dv/dt. This force is f = mr dω/dt, and its moment about the origin is N = mr^{2} d^{2}θ/dt^{2}, which is an angular form of Newton's Second Law. If we are dealing with a rigid body, then I = Σmr^{2} is called the *moment of inertia of mass*. We can then write N = I d^{2}θ/dt^{2}, with N analogous to force, and I analogous to mass. In the diagram, the applied moment N balances the accelerating moment. The forces do not have to have the same direction or magnitude, just the same moment.

In vectors, if **f** = d**p**/dt, then **N** = **r** x **f** = **r** x (d**p**/dt) = (d/dt)(**r** x **p**) = d**H**/dt, since **v** x **p** = 0 because **v** and **p** = m**v** are in the same direction. This shows that **N** = d**H**/dt in general.

So far we have only considered one coordinate system, in which Newton's Second Law holds. For this reason, it is called an *inertial system*. Newton's First Law defines what is meant by an inertial system in which the Second Law can be applied. Consider a second system, moving with uniform velocity **u** with respect to this system. If primes refer to this system, then **r**' = **r** + **u**t. Then, the velocity and acceleration in the moving system are **v**' = **v** + **u** and **a**' = **a**. Since the acceleration is the same, Newton's Law holds as well in the moving system. Systems moving with constant relative velocities to an inertial system are also inertial. This is the *principle of Galilean relativity*, which is valid so long as the relative velocities are much smaller than the speed of light. At greater velocities, we must use Einstein's principle of relativity instead.

Now let's consider a system rotating with a uniform angular velocity ω with respect to an inertial system. The diagram shows that in addition to the change in a vector in the rotating system, d**r**, the change **ω** x **r** must be added to get the full change δ**r** in the inertial system. This is a general result for the time derivatives of a vector, that can be applied to the velocity as well. We have **v**' = **v** - **ω** x **v**, and then **a**' = **a** - **ω** x (**ω** x **r**) - 2**ω** x **v**. Obviously, the rotating system is not inertial. However, it will seem like an inertial system if we add two *fictitious* forces to the acceleration, as defined by the second and third terms. The first is the *centrifugal force*, the second the *Coriolis* force. The centrifugal force is independent of the velocity and proportional to the square of the angular velocity, and is in a plane normal to the axis of rotation. The Coriolis force is proportional to the angular velocity and the linear velocity, and is normal to the plane defined by them.

A coordinate system fixed to the surface of the earth is such a rotating system. The angular velocity ω = 2π/86 164.1 s = 7.29211 x 10^{-5} radian/s. The sidereal (with respect to a fixed direction in space) period is used, not the period relative to the sun that we use for time. This is a small value, but has noticeable effects. As we mentioned above, the centrifugal force is included in the effective gravitational force m**g**, so the only fictitious force to be considered is the Coriolis force. The surface of the earth is normal to the direction of the effective gravity. Centrifugal force is the reason the earth is a spheroid. The effective component of the angular velocity is normal to a horizontal plane at the point of observation, with magnitude ω sin φ, where φ is the latitude. Its positive direction is upward in the Northern hemisphere, downward in the Southern hemisphere. The direction of the Coriolis force is normal to a horizontal velocity, so that it would turn the velocity clockwise in the Northern hemisphere, and anticlockwise in the Southern.

The Coriolis force has little effect on the water draining out of a bathtub, but a great effect on large-scale winds, where it causes winds to blow along isobars, instead of directly into a low-pressure area or out of a high-pressure area. The result is that winds circle a low-pressure area anticlockwise and a high-pressure area clockwise in the Northern hemisphere and that these areas persist as they move about. This is certainly the most visible effect of the Coriolis force, and additional proof that the earth rotates relative to an inertial system.

The Coriolis force also causes the plane of a pendulum to rotate clockwise in the Northern hemisphere, as can be seen by considering the velocity of the bob. The rate of this rotation is proportional to sin φ, so it vanishes at the equator and is a maximum at the poles. At the poles, the angular velocity is equal to that of the earth. At my latitude, 40°N, the pendulum makes one rotation in about 37 hours and 20 minutes. Foucault first observed this rotation, in 1851, in a famous experiment, and it is displayed in every good Science Museum. It is very difficult to create an unbiased support for the pendulum, and to release it properly, so it is a difficult home experiment. A ball dropped from a height in the Northern hemisphere winds up to the East of the point from which it was dropped (for a 200 ft drop, the distance is 0.31 inches).

Coulomb not only verified the inverse-square force between electric charges and measured them absolutely, but also established the laws of friction. Friction is an extremely variable phenomenon, but one of considerable practical interest to engineers. Coulomb's main result was that the force preventing two solids with plane faces in contact from sliding was independent of the area of contact, but proportional to the normal force pressing the two solids together. The ratio of the frictional force to the normal force is called the *coefficient of friction*, μ = F/N. Moreover, when the two solids were sliding, the coefficient was independent of the velocity of sliding. However, the *static* coefficient was somewhat greater than the *kinetic* coefficient. Frictional forces that obey these rules are called Coulomb friction. The coefficient of friction is dimensionless, the same in every system of units.

The coefficient of friction depends greatly on the preparation of the surfaces in contact. Clean, smooth surfaces will sometimes seize together in a vacuum so that they cannot be easily separated. It is never a good idea to have the same metal in contact at a bearing. For these reason, measurements of the coefficient of friction are not of great use, except to show average values. Nevertheless, we should have some idea of the range of values. Dry wood on wood shows μ = 0.25 - 0.50, wood on metal 0.2 - 0.6. Steel on steel is 0.58 dry, but falls to 0.16 when lubricated with light oil. Steel on brass, lubricated, gives 0.19.

The viscosity of a liquid or gas expresses the force necessary to move surfaces with respect to one another with the liquid or gas between them. It is assumed that the fluid adheres to the metal on either side, establishing a velocity gradient, as shown in the diagram. The viscosity is the ratio of the force per unit area in the direction of motion, divided by the velocity gradient, or η = (F/A)/(v/d) = Fd/Av. In cgs, the units are dyne-s/cm^{2}, given the name of poise, after Poiseuille. In fps, they are lb-s/ft^{2}. Using the known unit ratios dyne/lb and ft/cm, it is easy to find that 1 lb-s/ft^{2} = 478.8 poise. In fps, the lb-s/in^{2} or Reyn is often used, which is 3.325 poise. The viscosity of water at 20°C is 0.01 poise, or 1 centipoise, easy to remember. The viscosity of liquids is a strong function of temperature. In water, it varies from 1.787 cp at freezing to 0.2818 cp at boiling. Glycerine (sp.gr. 1.2613) has 1490 cp at 20°C, hexane (0.6603) 0.326 cp and methanol (0.7914) 0.597 cp. Light machine oil has 34.2 cp at 100°F (37.8°C). These figures give an idea of the viscosity of common liquids at normal temperatures.

The viscosity divided by the density is found in enough formulas that it has a name, *kinematic viscosity*, and a unit, *stokes*, of its own. Since the density of water is 1, its kinematic viscosity in stokes equals its viscosity in poise (but, of course, the units are different, and it is a different thing). The densities given above allow you to find the kinematic viscosity of a few liquids. Air, at 18°C, has a viscosity of 182.7 μpoise, which shows a typical value for a gas viscosity. Since its density is much lower than that of a liquid, its kinematic viscosity is not much different from that of a liquid (about 18 cp).

Hagen and Poiseuille studied the flow of water and blood, respectively, in capillary tubes around 1839-40. The flow through a tube of radius r and length s with a pressure difference of p dyne/cm^{2} between its ends is q = πpr^{4}/8sη cm^{3}/s, where η is the viscosity in poise. The *Reynolds number* R for this flow is R = DVρ/η, where D is the diameter of the pipe, and V the velocity of flow. Poiseuille's formula is valid when R is less than about 2000. Above this value, the flow becomes turbulent and different rules apply.

Another famous case of viscous flow is the movement of a sphere of radius a through a viscous fluid. In this case, the force resisting the motion is F = 6πηaV dynes, where η is in poise, a in cm, and V in cm/s. A glass marble 13 mm in diameter, falling in glycerine, will have a terminal velocity of V = (4π/3)(0.65)^{2} (981)(1.24)/6π(14.90) = 7.67 cm/s. This is found by equating the viscous force to the weight of the marble (sp.gr. = 2.5) less the buoyant force of the glycerine (sp.gr. 1.26). The Reynolds number in this flow is R = DVρ/η as with Poiseuille flow, and in this case is only 0.84, so viscous flow is guaranteed. Millikan used the same theory in his oil drop experiment to determine the charge on the electron. He measured the weight of a drop by the rate at which it fell in air.

Let's solve the pendulum with a small amplitude from what we have found above. This is motion in a plane with constant r, and the force on the bob of mass m is downward, and of amount mg. The moment of this force about the origin is N = mgl sin θ, tending to decrease θ, so the equation of motion is mgl sin θ = -ml^{2} d^{2}θ/dt^{2}. Note that the tension in the cord has no moment about the origin, and so does not enter into the equation of motion. For small oscillations, sin θ ≈ θ, so we have θ" + (g/l)θ = 0, where the primes indicate differentiation with respect to time. The solutions of this differential equation are well known, and θ oscillates sinusoidally with angular frequency √(g/l). A seconds pendulum, then, has a length l = g/π^{2} = 99.4 cm. The period of a seconds pendulum is 2 seconds, one second for each swing that releases the escapement one notch, so the clock ticks once a second. The pendulum was once used to measure g, but now the fall of a ball is measured electronically, to an accuracy of a milligal or better.

A ball leaves a cannon with a speed v_{0} of 1800 fps and an elevation of θ = 15°. How far will it go on a level field? The initial upward velocity is 466 fps, so the ball will reach the summit of its trajectory at a height h in 466/32.17 = 14.48 s, and fall back to earth in 28.97 s. In this time, it will travel horizontally d = 1739 fps x 28.97 s = 50,362 ft, or 9.54 miles. Of course, it will fall well short of this, because air resistance has a large effect, especially on supersonic cannonballs. If the muzzle velocity is only 180 fps, then it will travel for only 2.897 s, and plow up the ground only 504 ft away. This would be closer to correct, but it would be a really puny charge. Note that the range is not proportional to the muzzle velocity, but to its square.

Such simple problems as these could be multiplied indefinitely. They are not only excellent exercises, but also build confidence in your ability to understand at least this much of the universe.

speed of light c = 2.9979 x 10^{10} cm/s

electron charge e = 4.80325 x 10^{-10} esu

electron mass m = 0.9109534 x 10^{-27} g

Planck's constant h/2π = 1.05459 x 10^{-27} erg-s

Bohr magneton μ_{B} = 0.9274078 x 10^{-20} erg/gauss

Boltzmann's constant k = 1.38066 x 10^{-16} erg/K

universal gas constant R = 8.31441 x 10^{7} erg/K-mol

F. P. Beer and R. Johnston, Jr., *Vector Mechanics for Engineers*, 4th ed. (New York: McGraw-Hill, 1984). This is one of the best statics and dynamics texts for undergraduate engineers, but there are many others. There are many problems, and no advanced mathematics. This is the course, given to second-year students and lasting a year, that introduces students to vectors, scalar and vector products, and gives practice in differential and integral calculus. It reinforces good practice in solving problems, notably the free-body diagram. The mechanics part of the general physics course should do the same, but seems to make little impression on the average engineering student.

L. Page, *Introduction to Theoretical Physics*, 3rd ed. (New York: D. Van Nostrand, 1952). This text is mentioned because it is an excellent example of an advanced undergraduate review of classical physics, a "capstone" course that prepared the student for graduate study in physics. It covers vector analysis, particle dynamics, dynamics of rigid and deformable bodies, advanced dynamics (Lagrangian and Hamiltonian), hydrodynamics of perfect and viscous fluids, thermodynamics, statistical mechanics, kinetic theory of gases, electrostatics and magnetostatics, electric currents, electromagnetism, geometrical optics, physical optics and the origin of spectra. Each subject is treated rigorously and in serious depth, and the choice of subjects is wise. This book is probably long out of print, but should be remembered for its quality. It is an excellent reference and review. Similar, but more encyclopedic, is G. Joos, *Theoretical Physics*, 3rd ed. (New York: Dover, 1986) (reprint of the 1958 edition), which may be available.

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Composed by J. B. Calvert

Created 6 October 2002

Last revised 13 October 2007