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Scanned page · English
On floating bodies · p. 274
Converted · European Spanish

Las obras de Arquímedes

Entonces, puesto que CO = ½p, CL debe ser perpendicular a PT y, por tanto, a la superficie del fluido.

Y las fuerzas que actúan sobre la parte sumergida del paraboloide y sobre la parte situada por encima de la superficie actúan, respectivamente, hacia arriba y hacia abajo, a lo largo de rectas que pasan por F y H y son paralelas a CL.

Por tanto, el paraboloide no puede permanecer en la posición en que B apenas toca la superficie, sino que debe girar en el sentido de aumentar el ángulo PTM.

La demostración es la misma cuando el punto I no está en VP, sino en su prolongación, como en la segunda figura*.

Diagrama geométrico del original

[Con la notación de la p. 266, si la base BB′ toca la superficie del fluido en B, tenemos

BM=BVsinθ+PN

y, por la propiedad de la parábola,

BV2=(p+4AN)PV=pk(1+cot2θ)

Por tanto

ph=pk+p2cotθ

Para obtener el resultado de la proposición, debemos eliminar k entre esta ecuación y

x′−x=cosθ{p4(cot2θ+2)−23(h−k)}.

* Es curioso que las figuras de Torelli, Nizze y Heiberg sean todas incorrectas, pues sitúan el punto que he llamado I en BQ en vez de en la prolongación de VP.

274
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