2/33-pr-12
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| PROBLEMA XII, PROPOSITIO XXXIII. | PROBLEM XII, PROPOSITION XXXIII |
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| Dato perpendiculo et plano ad ipsum inclinato, quorum eadem sit altitudo idemque terminus sublimis, punctum in perpendiculo supra terminum communem reperire, ex quo si demittatur mobile, quod postea convertatur per planum inclinatum, ipsum planum conficiat tempore eodem, quo ipsum perpendiculum ex quiete conficeret. | Given a limited vertical line and an inclined plane of equal height, having a common upper terminal; it is required to find a point on the vertical line, extended upwards, from which a body will fall and, when deflected along the inclined plane, will traverse it in the same time-interval which is required for fall, from rest, through the given vertical height. |
| Sint perpendiculum et planum inclinatum, quorum eadem sit {20} altitudo, AB, AC: oportet, in perpendiculo BA, producto ex parte A, punctum reperire, ex quo descendens mobile conficiat spatium AC eodem tempore, quo conficit datum perpendiculum AB ex quiete in A. Ponatur DCE ad angulos rectos ad AC, et secetur CD aequalis AB, et iungatur AD: erit angulus ADC maior angulo CAD (est enim CA maior quam AB, seu CD). Fiat angulus DAE aequalis angulo ADE, et ad ipsam AE perpendicularis {255} sit EF, plano inclinato et utrinque extenso occurrens in F, et utraque AI, AG ponatur ipsi CF aequalis, et per G ducatur GH, horizonti aequidistans: dico, H esse punctum quod quaeritur. Intelligatur enim, tempus casus per perpendiculum AB esse AB; erit tempus per AC ex quiete in A ipsamet AC: {10} cumque in triangulo rectangulo AEF ab angulo recto E perpendicularis ad basim AF sit acta EC, erit AE media inter FA, AC, et CE media inter AC, CF, hoc est inter CA, AI: et cum ipsius AC tempus ex A sit AC, erit AE tempus totius AF, et EC tempus ipsius AI. Quia vero in triangulo aequicruri AED latus AE est aequale lateri ED, erit ED tempus per AF: et est EC tempus per AI: ergo CD, {20} hoc est AB, erit tempus per IF ex quiete in A: quod idem est ac si dicamus, AB esse tempus per AC ex G, seu ex H: quod erat faciendum. | Let AB be the given limited vertical line and AC an inclined plane having the same altitude. It is required to find on the vertical BA, extended above A, a point from which a falling body will traverse the distance AC in the same time which is spent in falling, from rest at A, through the given vertical line AB. Draw the line DCE at right angles to AC, and lay off CD equal to AB; also join the points A and D; then the angle ADC will be greater than the angle CAD, since the side CA is greater than either AB or CD. Make the angle DAE equal to the angle {255} ADE, and draw EF perpendicular to AE; then EF will cut the inclined plane, extended both ways, at F.Lay off AI and AG each equal to CF; through G draw the horizontal line GH. Then, I say, H is the point sought. For, if we agree to let the length AB represent the time of fall along the vertical AB, (Condition 2/02-th-02-cor2) then AC will likewise represent the time of descent from rest at A, along AC; and since, in the right-angled triangle AEF, the line EC has been drawn from the right angle at E perpendicular to the base AF, it follows that AE will be a mean proportional between FA and AC, while CE will be a mean proportional between AC and CF, that is between CA and AI. Now, since AC represents the time of descent from A along AC, (Condition 2/02-th-02-cor2) it follows that AE will be the time along the entire distance AF, and EC the time along AI. But since in the isosceles triangle AED the side EA is equal to the side ED it follows that ED will represent the time of fall along AF, while EC is the time of fall along AI. (Condition 2/11-th-11) Therefore CD, that is AB, will represent the time of fall, from rest at A, along IF; which is the same as saying that AB is the time of fall, from G or from H, along AC. |
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2/33-pr-12 |
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