Answer
(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 $1.19\times 10^5~Pa$
Work Step by Step
(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 $1.19\times 10^5~Pa$.
