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- 3: - 2hq y constant. Then c1 = # X Q is = Töö = a constant quantity = a, what it is at the vertex; that is, c1 is everywhere equal to KD. Consequently KR is = DP; and since RI is = PC, it is evident that KI is the same parabolic curve with DC, and may be placed any height above it, always producing an arch of equilibration, and very commodious for powder magazines.

PROMISCUOUS PROBLEMS, AS EXERCISES IN MECHANICS, STATICS, DYNAMICS, HYDROSTATICS, HYDRAULICS, PROJECTILES, &c. &c.

PROBLEM I,

Let AB and ac be two inclined planes, whose common altitude AD is given = 64 feet; and their lengths such, that a heavy body is 2 seconds of time longer in descending through Ab than through Ac, by the force of gravity; and if two balls, the one weighing 3 and the other 21b. be connected by a thread and laid on the planes, the thread sliding freely over the verter A, they will mutually sustain each other. Quere the lengths of the two planes 7

THE lengths of planes of the same height being as the times of descent down them (art. 195, Dynam.), and also as the weights of bodies mutually sustaining each other on them (art. 193), therefore the times must be as the weights: hence as 1, the difference of the weights, is to 2 sec. the diff. of times, : : } § : o: : the times of descending down the two planes. And as V16 : vá4::1 sec. : 2 sec. the time of descent down the perpendicular height. Then, by the laws of

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descents as 2 sec. : 64 feet :: } o: : #: : feet,the lengths of the planes. Note. In this solution we have considered 16 feet as the space freely descended by bodies in the 1st second of time, and 32 feet as the velocity acquired in that time, omitting the fractions or and #, to render the numeral calculations simpler, as was done in the preceding chapter on projectiles, and as we shall often do in solving the following questions, wherever great accuracy is not required.

Another Solution by means of Algebra.

Put z = the time of descent down the less plane; then

will z + 2 be that of the greater, by the question. Now the weights being as the lengths of the planes, and these again as the times, therefore as 2: 3: : z : z + 2 ; hence 2x + 4 = 3r, and z = 4 sec. Then the lengths of the planes are found as in the last proportion of the former solution.

PROBLEM II.

If an elastic ball fall from the height of 50 feet above the plane of the horizon, and impinge on the hard surface of a plane inclined to it in an angle of 15 degrees; it is required io find what part of the plane it must strike, so that after reflection, it may fall on the horizontal plane, at the greatest distance possible beyond the bottom of the inclined plane

** Here it is manifest that A.
the ball must strike the ob- I
lique plane continued on a IK E. G
point somewhere below the Se-P

horizontal plane; for other- F wise there could be no maxi- B N L. I mum. Therefore let Bc be M the inclined plane, cog the horizontal one, B the point on which the ball impinges after falling from the point A, BEGI the parabolic path, r its vertex, bh a tangent at B, being the direction in which the ball is reflected; and the other lines as are evident in the figure. Now, by the laws of reflection, the angle of incidence ABC, is equal to the angle of reflection ABH, and therefore this latter, as well as the former, is equal to the complement of the z- c, the inclination of the two planes; but the part IBM is = Z c, therefore the angle of projection HBI is = the comp. of double the Z c, and being the comp. of HBK, theref. 4. HBK = 2.4 c. Now, put

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Suppose a ship to sail from the Orkney Islands, in latitude 59° 3' north, on a N. N. E. course, at the rate of 10 miles an hour; it is required to determine how long it will be beJore she arrives at the pole, the distance she will have sailed, and the difference of longitude she will have made when she ar. rives there 2

Let ABC represent part of the equator; P the pole ; Amri' a loxodromic or rhumb line, or the path of the ship continued to the equator; PB, PC, any two meridians in. definitely near each other; nr, or mt, the part of a parallel of latitude intercepted between them.

Put c for the cosine, and t for the tangent of the course, or angle nmr to the radius r, Am, any variable part of the rhumb from the equator, = u ; the latitude Bm = w its sine w, and cosine y; and AB, the dif. of longitude from A, - z. Then, since the elementary triangle mnr may be considered as a rightangled plane triangle, it is, as rad, r ; c = sin. Z. mrn ; ; v

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ponding length of the rhumb ; then v =o ; and hence v –

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The same is evident without fluxions : for since the Zmrn is the same in whatever point of the path Amrp the point m is taken, each indefinitely small particle of Amrp, must be to the corresponding indefinitely small part of Bm, in the con. stant ratio of radius to the cosine of the course ; and therefore the whole lines, or any corresponding parts of them, must be in the same ratio also, as above determined. In the same manner it is proved that radius: sine of the course : ; distance: the departure.

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. ... tri: tri: property of the circle; theref # = Ty" TrāI. the general fluents of these are z = t X hyp. log. vote. which corrected by supposing z = 0, when r = a, are z = t x (hyp. log. v: -hyp. log. vo) ; but r x (hyp. log. wo: - hyp. log. vo is the meridional part of the dif orde latitudes whose sines are r and a, which call b ; then is 2: E #, the same as it is by Mercator's sailing.

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or converges to 0, which is its limit; consequently r is the limit or ultimate value of z ; but when x = r, the ship will be at the pole ; theref, the pole must be the limit, or evanescent state, of the rhumb or course : so that the ship may be said to arrive at the pole after making an infinite number 2r

m”-H 1 nishes when n, and consequently z, is infinite, in which case a is - r.

Now, from the equation D = o =#. it is found, that

of revolutions round it ; for the above expression wa

when d = 30°57' the comp. of the given lat. 59° 3' and c = sine of 67° 30' the comp. of the course, d will be = 2010 geographical miles, the required ultimate distance ; which, at the rate of 10 miles an hour, will be passed over in 201

hours or 84 days. The dif. of long, is shown above to be

infinite. When the ship has made one revolution, she will be but about a yard from the pole, considering her as a oint.

When the ship has arrived infinitely near the pole, she will go round in the manner of a top, with an infinite velocity; which at once accounts for this paradox, viz. that though she make an “infinite number of revolutions round the pole, yet her distance run will have an ultimate and definite value, as

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