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| PROBLEMA IX, PROPOSITIO XXIII. | PROBLEM IX, PROPOSITION XXIII |
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| Dato spatio, quovis tempore peracto ex quiete in perpendiculo, ex termino imo huius spatii planum inflectere, super quo post casum in {241} perpendiculo tempore eodem conficiatur spatium cuilibet spatio dato aequale, quod tamen maius sit quam duplum, minus vero quam triplum, spatii peracti in perpendiculo. | Given the time employed by a body in falling through a certain distance along a vertical line, it is required to pass through the lower terminus of this vertical fall, a plane so inclined that this body will, after its vertical fall, traverse on this plane, during a time-interval equal to that of the vertical fall, a distance equal to any assigned distance, pro{241}vided this assigned distance is more than twice and less than three times, the vertical fall. |
| Sit in perpendiculo AS tempore AC peractum spatium AC ex quiete in A, cuius IR maius sit quam duplum, minus vero quam triplum: oportet, ex termino C planum inflectere super quo mobile eodem tempore AC conficiat post casum per AC spatium ipsi IR aequale. Sint RN, NM {10} ipsi AC aequalia, et quam rationem habet residuum IM ad MN, eamdem habeat AC linea ad aliam, cui aequalis applicetur CE ex C ad horizontem AE, quae extendatur versus O, et accipiantur CF, FG, GO aequales ipsis RN, NM, MI: dico, tempus super inflexa CO post casum AC esse aequale tempori AC ex quiete in A. Cum enim sit ut OG ad GF, ita FC ad CE, erit, componendo, ut OF ad FG, seu FC, ita FE ad {20} EC, et ut unum antecedentium ad unum consequentium, ita omnia ad omnia, nempe tota OE ad EF, ut FE ad EC. Sunt itaque OE, EF, EC continue proportionales: quod cum positum sit, tempus per AC esse ut AC, erit CE tempus per EC, et EF tempus per totam EO, et reliquum CF per reliquam CO; est autem CF aequalis ipsi CA; ergo factum est quod fieri oportebat. Est enim tempus CA tempus casus per AC ex quiete in A, CF vero (quod aequatur CA) est tempus per CO post descensum per EC, seu post casum per AC: quod est propositum. Notandum autem est, quod idem accidet, si praecedens latio {242} non in perpendiculo fiat, sed in plano inclinato, ut in sequenti figura, in qua latio praecedens facta sit per planum inclinatum AS infra horizontem AE; et demonstratio est prorsus eadem. | Let AS be any vertical line, and let AC denote both the length of the vertical fall, from rest at A, and also the time required for this fall. Let IR be a distance more than twice and less than three times, AC. It is required to pass a plane through the point C so inclined that a body, after fall through AC, will, during the time AC, traverse a distance equal to IR.Lay off RN and NM each equal to AC. Through the point C, draw a plane CE meeting the horizontal, AE, at such a point that IM:MN = AC:CE. Extend the plane to O, and lay off CF, FG and GO equal to RN, NM, and MI respectively. Then, I say, the time along the inclined plane CO, after fall through AC, is equal to the time of fall, from rest at A, through AC.For since OG:GF = FC:CE, it follows, componendo, that OF:FG = OF:FC = FE:EC, and since an antecedent is to its consequent as the sum of the antecedents is to the sum of the consequents, we have OE:EF = EF:EC. Thus EF is a mean proportional between OE and EC. Having agreed to represent the time of fall through AC by the length AC (Condition 2/03-th-03-cor) it follows that EC will represent the time along EC, (Condition 202C) and EF the time along the entire distance EO, (Condition 2/11-th-11) while the difference CF will represent the time along the difference CO; but CF = CA; therefore the problem is solved. For the time CA is the time of fall, from rest at A, through CA while CF (which is equal to CA) is the time required to traverse CO after descent along EC or after fall through AC. Q. E. F.It is to be remarked also that the same solution holds if the antecedent motion takes place, not along a vertical, but along an inclined plane. This case is illustrated in the following figure where the antecedent motion is along the inclined plane AS {242} underneath the horizontal AE. The proof is identical with the preceding. |
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