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Unformatted text preview: 82. (a) We consider a thin slab of water with bottom area A and inﬁnitesimal thickness dh. We apply
Newton’s second law to the slab:
= (p + dp)A − pA =
=
= dFnet dp · A − dm · g
Adp − ρgAdh
dm · a = ρaAdh which gives
dp
= ρ ( g + a) .
dh
Integrating over the range (0, h), we get
h p= ρ(g + a)dh = ρh(g + a) .
0 (b) We reverse the direction of the acceleration, from that in part (a). This amounts to changing a to
−a. Thus,
p = ρ ( g − a) .
(c) In a free fall, we use the above equation with a = g , which gives p = 0. The internal pressure p
in the water totally disappears, because there is no force of interaction among diﬀerent portions of
the water in the bucket to make their acceleration diﬀerent from g . ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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