2/34-pr-13
|
|
||||
2/34-pr-13 |
|
|
|
|
|
| PROBLEMA XIII, PROPOSITIO XXXIV. | PROBLEM XIII, PROPOSITION XXXIV |
 | |
| Dato plano inclinato et perpendiculo, quorum idem sit sublimis terminus, punctum sublimius in perpendiculo extenso reperire, ex quo mobile decidens, et per planum inclinatum conversum, utrumque conficiat tempore eodem ac solum planum inclinatum ex quiete in eius superiori termino. | Given a limited inclined plane and a vertical line having their highest point in common, it is required to find a point in the vertical line extended such that a body will fall from it and then traverse the inclined plane in the same time which is required to traverse the inclined plane alone starting from rest at the top of said plane. |
| Sint planum inclinatum et perpendiculum AB, AC, quorum idem sit terminus A: oportet, in perpendiculo ad partes A extenso punctum sublime {256} reperire, ex quo mobile decidens et per planum AB conversum, partem assumptam perpendiculi et planum AB conficiat tempore eodem ac solum planum AB ex quiete in A. Sit horizontalis linea BC, et secetur AN aequalis AC; et ut AB ad BN, ita fiat AL ad LC; et ipsi AL ponatur aequalis AI, et ipsarum AC, BI tertia proportionalis sit CE, in perpendiculo AC producto signata: dico, CE esse spatium quaesitum, adeo ut, extenso perpendiculo supra A et assumpta parte AX ipsi CE aequali, mobile ex X conficiet utrumque spatium XAB aequali tempore ac solum AB ex A. Ponatur horizontalis XR aequidistans BC, cui occurrat BA extensa in R; deinde, producta AB in D, ducatur ED aequidistans CB, et supra AD {10} semicirculus describatur, et ex B ipsi DA perpendicularis erigatur BF usque ad circumferentiam: patet; FB esse mediam inter AB, BD, et ductam FA mediam inter DA, AB. Ponatur BS aequalis BI, et FH aequalis FB: et quia ut AB {257} ad BD, ita AC ad CE, estque BF media inter AB, BD, et BI media inter AC, CE, erit ut BA ad AC, ita FB ad BS; et cum sit ut BA ad AC, seu ad AN, ita FB ad BS, erit, per conversionem rationis, BF ad FS ut AB ad BN, hoc est AL ad LC. Rectangulum igitur sub FB, CL aequatur rectangulo sub AL, SF; hoc autem rectangulum AL, SF est excessus rectanguli sub AL, FB, seu AI, BF , super rectangulo AI, BS, seu AIB; rectangulum vero FB, LC est excessus rectanguli AC, BF super rectangulo AL, BF; rectangulum autem AC, BF aequatur rectangulo ABI (est enim ut BA ad AC, ita FB ad BI): excessus igitur rectanguli ABI super rectangulo AI, BF, seu AI, FH, aequatur {10} excessui rectanguli AI, FH super rectangulo AIB: ergo bina rectangula, AI, FH aequantur duobus ABI, AIB, nempe binis AIB cum quadrato BI. Commune sumatur quadratum AI: erunt bina rectangula AIB cum duobus quadratis AI, IB, nempe quadratum ipsum AB, aequale binis rectangulis, AI, FH cum quadrato AI. Communiter rursus assumpto quadrato BF, erunt duo quadrata AB, BF, nempe unicum quadratum AF, aequale binis rectangulis AI, FH cum duobus quadratis AI, FB, id est AI, FH. Verum idem quadratum AF aequale est binis rectangulis AHF cum duobus quadratis AH, HF; ergo bina rectangula AI, FH cum quadratis AI, FH aequalia sunt binis rectangulis AHF cum quadratis AH, HF; et dempto {20} communi quadrato HF, bina rectangula AI, FH cum quadrato AI erunt aequalia binis rectangulis AHF cum quadrato AH. Cumque rectangulorum omnium FH sit latus commune, erit linea AH aequalis lineae AI: si enim maior vel minor esset, rectangula quoque FHA et quadratum HA maiora vel minora essent rectangulis FH, IA et quadrato IA, contra id quod demonstratum est. Modo si intelligamus, tempus casus per AB esse ut AB, tempus per AC erit ut AC, et ipsa IB, media inter AC, CE, erit tempus per CE, seu per XA ex quiete in X: cumque inter DA, AB, seu RB, BA, media sit AF, inter vero AB, BD, id est RA, AB, media sit BF, cui aequatur FH, erit, ex praedemonstratis, excessus AH tempus per AB ex quiete in R, seu post {30} casum ex X, dum tempus eiusdem AB ex quiete in A fuerit AB. Tempus igitur per XA est IB; per AB vero post RA, seu post XA, est AI; ergo tempus per XAB erit ut AB, idem nempe cum tempore per solam AB ex quiete in A. Quod erat propositum. | Let AC and AB be an inclined plane and a vertical line respectively, having a common highest point at A. It is required to find a point in the vertical line, above A, such that a body, falling from it and afterwards having its motion directed along AB, will traverse both the assigned part of the vertical {256} line and the plane AB in the same time which is required for the plane AB alone, starting from rest at A. Draw BC a horizontal line and lay off AN equal to AC; choose the point L so that AB:BN = AL:LC, and lay off AI equal to AL; choose the point E such that CE, laid off on the vertical AC produced, will be a third proportional to AC and BI. Then, I say, CE is the distance sought; so that, if the vertical line is extended above A and if a portion AX is laid off equal to CE, then a body falling from X will traverse both the distances, XA and AB, in the same time as that required, when starting from A, to traverse AB alone. Draw XR parallel to BC and intersecting BA produced in R; next draw ED parallel to BC and meeting BA produced in D; on AD as diameter describe a semicircle; from B draw BF perpendicular to AD, and prolong it till it meets the circumference of the circle; evidently FB is a mean proportional between AB and BD, while FA is a mean proportional between DA and AB. Take BS equal to BI and FH equal to FB. Now since AB:BD = AC:CE and since BF is a mean proportional {257} between AB and BD, while BI is a mean proportional between AC and CE, it follows that BA:AC = FB:BS, and since BA:AC = BA:BN = FB:BS we shall have, convertendo, BF:FS = AB:BN = AL:LC. Consequently the rectangle formed by FB and CL is equal to the rectangle whose sides are AL and SF; moreover, this rectangle AL.SF is the excess of the rectangle AL.FB, or AI.BF, over the rectangle AI.BS, or AI.IB. But the rectangle FB.LC is the excess of the rectangle AC.BF over the rectangle AL.BF; and moreover the rectangle AC.BF is equal to the rectangle AB.BI since BA:AC = FB:BI; hence the excess of the rectangle AB.BI over the rectangle AI.BF, or AI.FH, is equal to the excess of the rectangle AI.FH over the rectangle AI.IB; therefore twice the rectangle AI.FH is equal to the sum of the rectangles AB.BI and AI.IB, or 2AI.FH = 2AI.IB+BI^2. Add AI^2 to each side, then 2AI.IB+BI^2+AI^2 = AB^2 = 2AI.FH+AI^2. Again add BF^2 to each side, then AB2+BF2 = AF^2 = 2AI.FH + AI2^ + BF^2 = 2AI.FH + AI^2 + FH^2. But AF^2 = 2AH.HF + AH^2 + HF^2; and hence 2AI.FH + AI^2 + FH^2 = 2AH.HF+AH^2+HF^2. Subtracting HF^2 from each side we have 2AI.FH+AI^2 = 2AH.HF+AH^2. Since now FH is a factor common to both rectangles, it follows that AH is equal to AI; for if AH were either greater or smaller than AI, then the two rectangles AH.HF plus the square of HA would be either larger or smaller than the two rectangles AI.FH plus the square of IA, a result which is contrary to what we have just demonstrated. If now we agree to represent the time of descent along AB by the length AB, (Condition 2/03-th-03-cor) then the time through AC will likewise be measured by AC; and (Condition 2/02-th-02-cor2) IB, which is a mean proportional between AC and CE, will represent the time through CE, or XA, from rest at X. Now, since AF is a mean proportional between DA and AB, or between RB and AB, and since BF, which is equal to FH, is a mean proportional between AB and BD, that is between AB and AR, (Condition 2/19-pr-06-cor) it follows, from a preceding proposition [Proposition XIX, corollary], that the difference AH represents the time of descent along AB either from rest at R or after fall from X, while the time of descent along AB, from rest at A, is measured by the length AB.(Condition 202C) But as has just been shown, the time of fall through XA is measured by IB, while the time of descent along AB, after fall, through RA or through XA, is IA.Therefore the time of descent through XA plus AB is measured by the length AB, which, of course, also measures the time of descent, from rest at A, along AB alone. Q. E. F. |
|
|
||||
2/34-pr-13 |
|
|
|
|
|