§2.1Διὰ τί, ἐὰν μὲν ἄνωθεν ᾖ τὸ σπαρτίον, ὅταν κάτωθεν ῥέψαντος ἀφέλῃ τὸ βάρος, πάλιν ἀναφέρεται τὸ ζυγόν, ἐἀν δὲ κάτωθεν ὑποστῇ, οὐκ ἀναφέρεται ἀλλὰ μένει;
Why is it that, if the cord is above, when, after the balance has inclined downwards, one removes the weight, the balance rises again, but if it is supported from below, it does not rise but remains?
Ἡ διότι ἄνωθεν μὲν τοῦ σπαρτίου ὄντος πλεῖον τοῦ ζυγοῦ γίνεται τὸ ἐπέκεινα τῆς καθέτου;
Is it because, when the cord is above, the part of the balance beyond the perpendicular becomes greater?
Τὸ γὰρ σπαρτίον ἐστὶ κάθετος.
For the cord is a perpendicular.
§2.2Ὤστε ἀνάγκη ἐστὶ κάτω ῥέπεν τὸ πλέον, ἕως ἂν ἔλθῃ ἡ δίχα διαιροῦσα τὸ ζυγὸν ἐπὶ τὴν κάθετον αὐτήν, ἐπικειμένου τοῦ βάρους ἐν τῷ ἀνεσπασμένῳ μορίῳ τοῦ ζυγοῦ, Ἕστω ζυγὸν ὀρθὸν ἐφ’ οὖ ΒΓ, σπαρτίον δὲ τὸΑΔ. Ἐκβαλλόμενον δὴ τοῦτο κάτω κάθετος ἔσται ἐφ’ ἦς ἡ ΑΔΜ.
Thus it is necessary that the greater part inclines downwards until the line bisecting the balance comes upon the perpendicular itself, while the weight is placed upon the raised part of the balance. Let there be a horizontal balance BC, and let the cord be AD. If this is produced downwards, it will be the perpendicular ADM.
§2.3Ἔὰν οὖν ἐπὶ τὸ Β ἡ ῥοπὴ ἐπιτεθῇ, ἔσται τὸ μὲν Β οὖ τὸ Ε, τὸ δὲ Γ οὖ τὸ Ζ ὥστε ἡ δίχα διαιροῦσα τὸ ζυγὸν πρῶτον μὲν ἦν ἡ ΔΜ τῆς καθέτου αὐτῆς, ἐπικειμένης δὲ τῆς ῥοπῆς ἔσται ἡ ΔΘ· ὥστε τοῦ ζυγοῦ ἐφ’ ᾦ ΕΖ τὸ ἔξω τῆς καθέτου τῆς ἐφ’ ἦς ΑΒ, τοῦ ἐν ὦ ΦΠ, μείζω τοῦ μίσεος.
If then the weight is placed upon B, B will be where E is, and C where Z is; so that the line bisecting the balance, which at first was DM, the perpendicular itself, will be DTh when the weight is applied; so that of the balance EZ, the part outside the perpendicular AB, namely that in PhP, is greater than half.
§2.4Ἐὰν οὖν ἀφαιρεθῇ τὸ βάρος ἀπὸ τοῦ Ε, ἀνάγκη κάτω φέρεσθαι τὸ Ζ ἔλαττον γάρ ἐστι τὸΕ. Ἐὰν μὲν οὖν ἄνω τὸ σπαρτίον ἔχῃ, πάλιν διὰ τοῦτο ἀναφέρεται τὸ ζυγόν.
If then the weight is removed from E, Z must be carried downwards; for E is less. If then the balance has the cord above, for this reason it rises again.
Ἐὰν δὲ κάτωθεν ᾖ τὸ ὑποκείμενον, τοὐναντίον ποιεῖ· πλεῖον γὰρ γίνεται τοῦ ἡμίσεος τοῦ ζυγοῦ τὸ κάτω μέρος ἢ ὡς ἡ κάθετος διαιρεῖ ὥστε οὐκ ἀναφέρεται· κουφότερον γὰρ τὸ ἐπηρτημένον.
But if the supporting part is below, it does the opposite; for the lower part becomes more than half of the balance as divided by the perpendicular, so that it does not rise; for the suspended part is lighter.
§2.5Ἕστω ζυγὸν τὸ ἐφ’ οὖΝΞ, τὸ ὀρθόν, κάθετος δὲ ἡ Κ Λ Μ. Δίχαδὴ διαίρεῖται τὸΝΞ. Ἐπιτεθέντος δὲ βάρους ἐπὶ τὸ Ν, ἔσται τὸ μὲν Ν οὖ τὸ Ο, τὸ δὲ Ξ οὖ τὸ Ρ, ὁ δὲ Κ Λ οὖ τὸ ΑΘ, ὥστε μεῖζόν ἐστι τὸ ΚΟ τοῦ Λ Ρ τῷ ΘΚΛ. Καὶ ἀφαιρεθέντος οὖν τοῦ βάρους ἀνάγκη μένειν· ἐπίκενται γὰρ ὥσπερ βάρος ἡ ὑπεροχὴ ἡ τοῦ ἡμίσεος τοῦ ἐν ᾦτὸ Κ.
Let there be a horizontal balance NX, and the perpendicular KLM. NX is bisected. When a weight is placed upon N, N will be where O is, and X where R is, and KL where ATh is; so that KO is greater than LR by ThKL. Therefore, even when the weight is removed, it must remain; for the excess of the half in which K is lies upon it like a weight.