Asked by Ana
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 x 10^5 N/m^2, the density of water is 1 g/cm^3, g=9.8m/s^2.)
(Answer: 1.44 x 10^6 N)
(Answer: 1.44 x 10^6 N)
Answers
Answered by
Damon
using Newtons, kg, meters
water pressure = 1000 kg/m^3*9.8 m/s^2 * 1.5 m = 14,700 N/m^2
so total pressure = 14,700 +100,000 = 114,700 N/m^2
area = pi d^2/4 = pi 4^2/4 = 12.6 m^2
s total force down = 114,700 *12.6
= 1441400 = 1.44*10^6 sure enough
Note, the air also pushes up on the bottom of the pool so things are not as bad as they look
water pressure = 1000 kg/m^3*9.8 m/s^2 * 1.5 m = 14,700 N/m^2
so total pressure = 14,700 +100,000 = 114,700 N/m^2
area = pi d^2/4 = pi 4^2/4 = 12.6 m^2
s total force down = 114,700 *12.6
= 1441400 = 1.44*10^6 sure enough
Note, the air also pushes up on the bottom of the pool so things are not as bad as they look
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