## Physics: Principles with Applications (7th Edition)

(a) The absolute pressure at the bottom of the pool is $1.19\times 10^5~Pa$ The total force on the bottom of the pool is 500 N (b) Since the pressure acts in all directions, the pressure against the side of the pool near the bottom is also $1.19\times 10^5~Pa$
(a) We can find the absolute pressure at the bottom of the pool. $P = 1~atm+\rho~g~h$ $P = 1.01\times 10^5~Pa +(1.00\times 10^3~kg/m^3)(9.80~m/s^2)(1.8~m)$ $P = 1.19\times 10^5~Pa$ The absolute pressure at the bottom of the pool is $1.19\times 10^5~Pa$ We can find the total force on the bottom of the pool. $F = P*A$ $F = {(1.19\times 10^5~Pa)}{(28.0~m)(8.5~m)}$ $F = 2.8*10^7~N$ The total force on the bottom of the pool is 2.8*10^7 N. (b) Since the pressure acts in all directions, the pressure against the side of the pool near the bottom is also $1.19\times 10^5~Pa$.