P15_082

# P15_082 - 82. (a) We consider a thin slab of water with...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 . ...
View Full Document

## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online