unit of length, and inversely as the unit of time. This is usually expressed by saying that the dimensions of the unit of velocity are of one degree in length, and minus one degree in time; or that the dimensions of velocity are or LT1.

T

It should be noticed that is not a number, or a ratio in the strict Euclidean sense. Nor does the symbol indicate division in the usual sense; for we cannot divide a length by a time. The symbol rather indicates the idea which is implied in our usual use of the word per; as when we speak of a velocity of 30 feet per second. When, therefore, we write

we mean that the unit of velocity (V) is such that the space L is described per time T.

8. Acceleration.-Proceeding with this reasoning we shall find that in acceleration time is involved twice. For acceleration is measured by the increase of velocity per unit time; so that if A denote the unit of acceleration, A is equal to V per T, or =V/T, and since we have already seen that

We can arrive at the same result more formally as follows:

Let A,V,L, and T represent respectively the units of acceleration, velocity, length and time in one system, A',V',L', and T' the corresponding quantities in a second system; and let a denote an acceleration such that the

velocity v is generated in the time t. of the quantity in the first system,

n =

a

Α ́

If n be the measure

But the measure of an acceleration is the number of units of velocity generated per unit of time, so that

But since the quantity measured in both systems is the same, we have,

A comparison of equations (5) and (6) shows that the unit of acceleration varies directly as the unit of length and inversely as the square of the unit of time; in other 늘 or LT-2

words, that its dimensions are

The following example will illustrate the way in which these equations are applied

Ex. 1. Express the acceleration due to gravity in terms of the mile and the hour as the units of length and time, its value being 32 when the foot is the unit of length, and the second the unit of time.

an equation which gives us the required measure (n') in the new system, when the relations between the fundamental units in the old and new systems are known. In the example n = = 32,

9. Force, Work, and Power.-Before proceeding to give the dimensions of other derived units in mechanics, it may be well to point out the considerations which determine the choice of any new unit based upon the fundamental quantities or upon derived units which have already been fixed. We shall take the unit of force as our example.

According to the second law of motion, force is measured by the change of momentum which it pro

duces, i.e.

fx (rate of change of mv),

... fx ma,

(where a denotes acceleration), or

f=k'ma.

The units of mass and acceleration are already fixed, but we may make the unit of force whatever we please, and it will obviously be most convenient to choose it so that the constant multiplier k shall be equal to unity. Our equation will now become

f=ma.

Now suppose m and a to be each equal to unity; then ƒ will also be equal to unity. Thus our unit of force (F) is defined as being that force which produces unit acceleration in unit mass. We may therefore write

Work is measured by the product of force into the distance through which the force acts. Hence the dimensions of work will be those of force multiplied by length, or

W=MLT-2xL=ML2T-2,

The power (or activity) of an agent is measured by the rate at which it does work; hence the dimensions of power are

10. Change of Units.— Knowing the dimensions of these quantities, we can perform the change of units without going through the lengthy reasoning of §§ 7 and 8; we shall indicate the general method to be followed, but it will be best understood by reference to the actual examples given. [See also equations (5) and (6) in $ 8.]

Let 9 be any concrete quantity, and let its measure be n in terms of the unit Q, which is based upon the fundamental units M, L, and T; we wish to find its measure n' in a new system in terms of the unit Q' which is based upon M', L', and T'. Since the quantity measured in both systems is the same,

Let the dimensions of Q be M*L*T*; substituting for Q and Q' in equation (7) we have

Ex. 2. Find the number of dynes in a poundal (the poundal being the British absolute unit of force, based upon the pound, foot, and second).

Referring to equations (7) and (8) we see that, since n = 1, the required number n' is the change-ratio or multiplier for changing from British to G.C.S. units of force. The dimensions of force are MLT-2, so that x=1, y=1, and z=−2. T and T', the units of time, are the same (one second) in both systems.

Ex. 3. Find the value of a horse-power in watts, a horse-power being equivalent to 550 foot-pounds per second, and the value of g being 32.18.

As the foot-pound is a gravitation unit, we shall first have to reduce to the corresponding absolute unit by multiplying by g

550 foot-pounds = 550 × 32.18 foot-poundals.

The dimensional equation for finding the equivalent rate of working in G.C.S. units (ergs per second) is,

550 × 32.18 × ML2T3=n' × M'L'2T'-3.