Balance beam a m n o p: reducing weights to the pole
Using earlier conclusions to find the load that balances the beam
A single worked problem, with a beam labelled a m n o p and a hanging load, applies Leonardo's earlier balance conclusions to reduce distributed weights to points near the pole. The weight of 6 at a n is taken to act wholly at m and the weight of 2 at n p at o; since n o goes three times into m n, a weight at o balances a third of itself at m. From this he finds that 6 at m equals 12 at o, so hanging a further 10 makes the required 12.
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Reducing arm weights to points m and o
By the first conclusion the weight of 6 at a n acts entirely at m, and the weight of 2 at n p at o; by the third, n o goes 3 times into m n. Therefore 6 at m stands equal with 12 at o, so hanging 10 of weight, together with the 2 at n p, makes 12.
A subtriple distance and its balancing weight
Because n o is a subtriple of m n, every weight at o stands in balance with a third of itself, a subtriple, at m. Thus, there being 2 at o, it lifts two thirds of a whole unit at m.