What is the total force on the bottom of a 400-cm-diameter by 1.5-m-deep round wading pool due to the weight of the air and the weight of the water? (Note the pressure contribution from the atmosphere is 1.0 × 105 N/m2, the density of water is 1 g/cm3, and g = 9.8 m/s2.)

Answer 1

Water density = 1000 kg/m^3 which is 1 g/cm^3
Pa = 10^5 N/m^2 Pacals or N/m^2
additional pressure = rho g h = 1000 * 9.8 *1.5 = 9,800*1.5 = 14,700 Pascals
Total pressure down = 100,000 +14,700 = 114,700 Pascals
radius R = (400/100) /2 = 2 meters
pi R^2 = 3.14 * 4 = 12.56 m^3
total force down = 114,700 * 12.56 = 1,449,632 Newtons
(Remember that 100,000 Pascals of atmospheric also pushes up on the bottom from underneath. It is the 14,700 additional from the water that matters for design.)