#### 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.
$F = \frac{P}{A}$
$F = \frac{1.19\times 10^5~Pa}{(28.0~m)(8.5~m)}$
$F = 500~N$
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 $1.19\times 10^5~Pa$.