Bukhovtsev-et-al-Problems-in-Elementary-Physics

MECHANICS
257
Fig. 381 Fig. 382
252. A liquid moves in a syphon by the action of the forces of cohesion
between the elements of the liquid. The liquid in the long elbow outweighs
the liquid in the short one, and pumps it over. It could be assumed on this
basis that water can be pumped over a wall of any height with the ski of
a syphon. This is not so, however. At a lifting height of 10 metres the
pressure inside the liquid becomes zero. The air bubbles always present in
water will begin to expand, and the water column will be broken, thus
stopping the action of the syphon.
253. The device will first act as a syphon and the water will flow through
the narrow pipe into the reservoir. Then an air bubble will slip through A
and divide the liquid in the upper part into two portions. After this the
liquid will no longer flow out. .
254. The pressure of the water directly under the piston of each pump
is less than atmospheric pressure by pg (H +h), where p is the density of
water. Therefore, to keep the piston in equilibrium, it should be pulled
upward with the force F = pg (H+h) A, where A is the area of the piston.
Hence, a greater force should be applied to the pistons with a greater area.
255. The pressure on the bottom is P=pgbH+h) (Fig. 382). On the
other hand, since the vessel is a cylinder, p= :R~.
The height h can be determined if we equate the forces acting on the piston
pghn (RI-r 2 )= G
Hence,
1 ( G r 2 )
nR2p g R2~r2
H=-- m----- ~ 10 cm
256. To prevent flowing out of the water, the vessel should be given such
an acceleration at which the surface of the water takes the position shown
in Fig. 383. The maximum volume of the water is ~: and the mass of the
entire system is M+be'; p. The required acceleration can be found from the
condition that the sum of the forces acting on a small element of the water
with a mass 11m near the surface is directed horizontally (Fig. 383).
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