a body, the momentum produced in a given time is proportional to the force. This law enables us to answer the question, if a force of P lbs. acts on a body containing м lbs. of matter, and in each second accelerates its motion by f feet per second, what is the relation between P, M, and f? In this case the force in one second increases the momentum of the body by мƒ; but if a mass of 1 lb. fall freely at London it is acted on by a force of 1 lb., and in each second its velocity is increased by 32.1912 ft. per second; that is, its momentum is increased in one second by 1 x 32.1912. Hence

The constant divisor 32.1912 in this equation can be got rid of by using as the unit of mass 32.1912 lbs. of matter. Now, as a pound of distilled water at the standard temperature contains 27-7274 cubic inches, 32-1912 lbs. of water contains 892.6 cubic inches. Hence we have

where the unit of force in which P is estimated is pounds, the units of space and time, feet and seconds, and the unit of mass, the quantity of matter in 892.6 cubic inches of distilled water at the standard temperature.

To express the same relation it is, however, much more convenient to adopt a different unit of force. It is plain that the third law of motion amounts to this: we may consider that force to be unity which, by acting on the unit of mass for a unit of time, produces the unit of velocity. If, then, the mass of the body be м lbs. of matter, and if the force by acting on it for one second produces a velocity of ƒ feet per second, the number of units of force in the force will be мƒ or

When force is thus measured it is said to be measured in (British) absolute units.

A comparison of equations (1) and (3) makes it plain that the pound (of force) is equal to 32-1912 (British) absolute units of force.

If we suppose a mass of м lbs. to be at a place where the accelerative effect of gravity is g', the force of gravity on that body at that place in absolute units will be мg', and therefore in pounds it will be мg'÷g, where g denotes 32.1912.

If a mass м is moved from rest by a constant force P, and if ƒ is the accelerative effect of P on м, we have

If now м has acquired a velocity v after moving from rest over a distance h, we have

Therefore

2fh=v2 Ph=Mv2

(5)

(6)

In like manner if p's action had changed м's velocity from v to v, we should have had

Now, referring to equation (6), let us suppose that v is measured in feet per second, and P in pounds, then h will be in feet, and consequently Ph will denote the units of work done by P, and accumulated in м; it follows from equation (4) that for (6) to be true м must be estimated in units such that 892.6 oz. of distilled water is the unit of mass. If м is estimated in pounds, the expression

MV2

must be divided by 32-1912 or approximately by 32 to give the units of work accumulated (compare Art. 119. A similar remark applies to equation (7). The expres

sion MV2 is commonly called the vis viva, of a particle whose mass is м and velocity v.

In the following examples, which illustrate the subject of Arts. 131, 132, the numerical value of g in London is taken strictly as 32-1912, in other parts of the book the approximate value 32 is used except where the contrary is stated.*

Ex. 711.-A force estimated in British absolute units is 90; what would be the numerical value of the force if the unit of mass were 1 units of space and time a yard and a minute?

cwt. and the Ans. 964.3.

Ex. 712.-How many British absolute units would there be in an absolute unit of force if the kilogramme, mètre, and second were employed as the units of mass, space, and time? A kilogramme equals 2.2055 lbs. and a mètre equals 39.3708 inches. Ans. 7.236.

Ex. 713.-A mass of 500 lbs. is acted on by a force of 125 absolute units; what space will it describe from rest in 8 seconds? Ans. 8 ft.

Ex. 714.-A cubic foot of cast iron is observed to increase its velocity by 5 ft. per second in every second of its motion; determine the force which produces this acceleration, assuming that a cubic foot of water weighs 1,000 oz. Ans. (1) 2252.2 absolute units. (2) 69.96 lbs.

Ex. 715.-A body whose mass is 10 lbs. moves from rest along a straight line under the action of a constant force; during the first second of its motion it describes a distance of 50 ft. What is the force which produces the motion? What is the mass of the body which that force would just support in London? What in Trinidad? (Table XV). Ans. (1) 1000 absolute units. Ex. 716.-The accelerative effect of the moon's attraction on a point situated on its surface is about 5.4. A man can jump to a height of 5 feet on the earth's surface; how high could he jump on the moon's surface? Ans. 29.6 ft.

(2) 31.06 lbs. (3) 31-16 lbs.

[In both cases the mass of the man's body is the same, and the force exerted by the muscles is the same, quite irrespectively of the force of gravity; consequently, in the act of jumping, the velocity with which he leaves the ground is the same in both cases.]

Ex. 717. Suppose that near the surface of the moon a mass of 5 lbs. were placed on a plane, and it were observed that the plane descended with a uniform acceleration of 3.6 ft. per second; what pressure would the mass

* The pound has been used as the unit of mass throughout the remaining pages of the book. This accounts for the answers to several of the questions differing from those given in previous editions.

Herschel's Outlines of Astronomy, Art. 508.

produce on the plane in absolute units? How much matter would a force equal to this pressure support in London? Ans. (1) 9. (2) 4.47 oz.

Ex. 718.-If near the moon's surface two bodies are connected by a fine thread passing over a smooth fixed cylinder, the former, containing 30 lbs. of matter, falls and pulls up the latter, which contains 20 lbs. of matter; find the tension of the thread (1) in absolute units (2) in pounds. Ans. (1) 129 6. (2) 4.02. Ex. 719.-A mass of 7 lbs. falls through 100 ft. near the surface of the moon; how many units of work does it accumulate in the fall?

Ans. 117.4. Ex. 720.-If equal forces (P) act on two unequal bodies for the same time, show that the bodies will acquire equal momenta.

Ex. 721.-Show that momenta of the bodies in the last Example will be equal when P varies from instant to instant, provided the forces are the same at the same instant throughout their time of action.

[The results in the last two Examples are of considerable importance; they are almost self-evident and therefore liable to be forgotten-for this reason the student's attention is particularly directed to them.]

Ex. 722.-When the powder in the bore of a cannon is exploded, the pressures on the end of the bore and on the shot, are at each instant equal: a shot weighing 6 lbs. is fired from a gun quite free to move and weighing 6 cwt.; the velocity with which the shot leaves the gun is 1000 ft. per second; what is the velocity of the gun's recoil? Ans. 8.93 ft. per sec.

Ex. 723.-Show that the number of units of work accumulated in the gun is always small compared with the number accumulated in the shot; and ascertain these numbers in the case suggested in the last Example. Ans. 93,750 units in the shot and 837 in the gun.

Ex. 724.-What reason can be assigned for the practical rule that, cæteris paribus, the velocity of the shot is proportional to the square root of the weight of the charge?

Ex. 725.-If the trunnions of the gun in Ex. 722 are supported on two parallel smooth planes inclined at an angle of 30°, determine how far it will move along these planes. Ans. 2.5 ft.

Ex. 726.—The mercurial barometer (when placed on the sea-level) stands at any given height; suppose the force of gravity to undergo any change either of increase or decrease; explain why this, unaccompanied by any other change, would not alter the height of the barometer.

Ex. 727.-There are two mercurial barometers placed on the sea-level; the height of the mercury in the one is 30 in., and there is a perfect vacuum above it; the height of the mercury in the other is 28 in., and there is an imperfect vacuum of 6 in. above it; suppose the force of gravity to change from 32 to 24, at what height will the mercury now stand in the second barometer? Ans. 32-25 in.

CHAPTER IV.

THE CONSTRAINED MOTION OF A POINT.

Proposition 27.

The velocity acquired by a body in sliding from one point to another on a smooth curve is the same as that acquired by a body which falls freely through a space equal to the vertical height of the higher above the lower point.

A

FIG. 163.

A

Let A and B be the two points, draw BB' horizontal and AB' vertical; let the body leave a with a velocity v, and arrive at в with a velocity v; then, if м be the mass of the body, the number of units of work accumulated in it while it moves from A to B will equal (Art. 132)

≥ M (v2 — v2)

B

B'

Now, the only forces that have acted on the body are its weight and the reaction of the curve; the work done by the former of these equals w× A B′ or Mg X A B', and the latter does no work, since its direction is always perpendicular to that in which its point of application is moving (Art. 103); therefore

M (v2 —v2)=WX A B'

... v2-v2=2g x A B′

But this equation likewise gives the velocity (v) of a body supposed to leave a with a velocity v, and to fall freely to B'. Therefore, &c.

Q. E. D.