the action of one pair of teeth, and the same being true of every pair of teeth we obtain the result stated in the question. Of course, the addition of the expression contained in the present question to that obtained in the last is the correct approximate formula for the work expended in raising a weight through the intervention of a pair of toothed wheels.]

Ex. 559.-A force P acting at the end of an arm o A, two feet long, causes the toothed wheel oв to make 10 turns per minute; this wheel working with the wheel oв turns the drum o1c and raises the weight q; given that P does at the point ▲ 330,000 units of work per minute, determine approximately the weight o that will be raised by the drum, having given the radius of Oв to be 1 foot, o1в to be 3 feet, the number of teeth in OB to be 40, and the radius of the drum 5 feet; the teeth, axles, and bearing are all of cast iron without unguents; the radii of the axles are 3 in., the weight of the axles and appendages of o is 3600 lbs., and that of o1 is 5400 lbs. Ans. 2752 lbs.

[See Note to Ex. 557.]

Ex. 560.--Show that in a train of p pairs of wheels and pinions* the work lost by friction between the teeth is given by the formula

where ni, na, nз and pinions.

. nap are the number of teeth in the successive wheels

Ex. 561.-There is a train of p equal pairs of wheels and pinions; the numbers of teeth are such that the last axle revolves m times faster than the first; show that if u is the number of units of useful work yielded, the work lost by the friction between the teeth is represented by the formula

where n is the number of teeth in each wheel.

*Ex. 562.—If it is required to make the last axle move m times faster than the first, show that the loss of work is least when p, the number of pairs of wheels and pinions is given by the formula

*Ex. 563.-If in the last Example it is required to multiply the velocity 100 times, show that the proper number of pairs of wheels and pinions is 3 or 4, i.e. show that the equation in the last Example gives a value of P between 3 and 4; and determine the number of teeth employed in each case if the first pinion have 20 teeth, using the nearest whole numbers.

Ans. (1) 339. (2) 333.

* When a small wheel drives a large one the former is frequently called

a pinion and the latter a wheel.

2π

n

Ex. 564.—If in the pair of wheels already described (Art. 109) all but a single pair of teeth be cut away, so that the remaining teeth act on each other while the wheel o moves through an angle- before coming to the line of centres, and also while it moves through an equal angle after having passed the line of centres, and if we suppose P and Q to act on the pitch circles of their respective wheels, show that when the point of contact is in such a position that the wheel o has to revolve through an angle the point of contact comes to the line of centres we have

before

and that when the point of contact is so situated that the driver has revolved through an angle 0 from the line of centres we have

[If in the accompanying figure x is the point of contact of the teeth before they come to the line of centres, that point x will be on the circumference

of a circle whose diameter is oa; if then we draw a line RR' such that the angle RXA equals o, this will be the line of the mutual action of the teeth; remembering that the angle Aox equals @ it is easily shown that the perpendiculars on RR' from O and o1 are respectively equal to

and

r cos e cos

(r+r1) cos (0 + 4) –

-r cos e cos

whence the first equation is obtained; the second is obtained in a similar manner, by determining the relation between P and Q when the follower has revolved through an angle e' which will be found to be

pr=Q{r+(r+r1) tan
tan ở tan

whence we obtain the answer.]

Ex. 565.—If AB be any diameter of a circle APB; if c be any point taken in the prolongation of AB (so that в is between a and c), and if AP, BP, CP be joined, show that

BC AC tan PAB tan BPC

and hence explain the action of the forces which produces the result that follows from the first equation in Ex. 564, viz. that when r1 =(r+r1) tan 0 tan & the force P must be infinitely large to bring a into the state bordering on motion.

Ex. 566.—If the driver be not greater than the follower, show from the equations of Ex. 564, that for a given value of a, the value of P is greater when the driving tooth is in a given position before it comes to the line of centres than when it is in a corresponding position after having passed the line of centres.

[If m be written for the ratio of r to r1 (so that m cannot be greater than unity) the equations in Ex. 564 can be written thus:

and

consequently

P-P=Q

ms}

{I+m./
1 + m.utane - (1 + 1)μ u tan me+positive terms

m

and this, on expanding in power of 0, is found to equal

μα

{1+m.

· 0 ( 1 − m2 02 + 1/2/3 1-m101+ .)+ positive terms]

PART II.

DYNAMICS.

CHAPTER I.

INTRODUCTORY.*

111. Velocity. Before considering force as the cause of velocity or of change of velocity, it will be necessary to define accurately the means of estimating the magnitude of velocities.

Def.-A body moves uniformly or with a uniform velocity when it passes over equal spaces in equal times.

The units of space and time commonly employed are feet and seconds:† and whenever a body is said to be moving with any particular velocity, e.g. 5 or 6, this will always mean with a velocity of 5 or 6 feet per second.

Def. When a body moves with a variable velocity, that velocity is measured at any instant by the number of units of space it would pass over in a unit of time if it continued to move uniformly from that instant.

It will be seen from the definition that variable velocity is measured in a manner that exactly falls in with the

*The student is particularly recommended to make himself thoroughly master of this chapter before proceeding further.

To prevent mistake, it may be stated that the time referred to is mean solar time.

ordinary way of speaking: thus, when we say that a train is moving at the rate of 40 miles an hour, we mean that if it were to keep on moving uniformly for an hour, it would pass over 40 miles. Again, if we were to drop a small heavy body, we should find that at the end of a second it is moving at the rate of about 32 feet per second, or, as it is commonly stated, it acquires in a second a velocity 32, meaning that if it were to move uniformly from the end of that second it would pass over 32 feet in each successive second.

112. Relation between uniform Velocity, Time, and Space. In the case of a body moving with a uniform velocity, it is evident that the number of feet (8) passed over in t seconds must be t times the number of feet passed over in one second (v),

..s=vt.

The space 8 can, of course, be represented geometrically by the area of a rectangle whose sides severally represent on the same scale the velocity and the time.

Ex. 567.-A body moves uniformly over 2 miles in half an hour, determine its velocity. Ans. 7.

2.5280
3600

Ex. 568.-A body moves at the rate of 12 miles an hour, determine its velocity. Ans. 17. Ex. 569.-The equatorial diameter of the earth is 41,847,000 ft., and the earth makes one revolution in 86,164 seconds: determine the velocity of a point on the earth's equator. Ans. 1526.

Ex. 570.-A body moves with a velocity of 12; how many miles will it pass over in one hour? What would be its velocity if we used yards and minutes as units instead of feet and seconds? Ans. (1) 8. (2) 240.

113. The Velocity acquired by Falling Bodies.-It appears as the result of the most careful experiment that at any given point of the earth's surface, a body falling freely in vacuo acquires at the end of every second at