Question

You are assigned the design of a cylindrical, pressurized water tank for a future colony on Mars, where the acceleration due to gravity is 3.71 m/s^2. The pressure at the surface of the water will be 100 kPa , and the depth of the water will be 14.1 m . The pressure of the air outside the tank, which is elevated above the ground, will be 92.0 kPa . Part APart complete Find the net downward force on the tank's flat bottom, of area 2.00 m^2 , exerted by the water and air inside the tank and the air outside the tank. Assume that the density of water is 1.00 g/cm^3.

Answer 1

Answer:

488.6KN

Explanation:

Hello!

the first step to solve this problem we must find the pressure exerted at the bottom of the tank (P) which is the sum of the external air pressure (P1 = 92kPa), the pressure inside the tank (P2 = 100kPa) and the pressure due to the weight of the water (P3), taking into account the above we have the following equation

P=P1+P2+P3

to find the pressure at the bottom of the tank due to the weight of the water we use the following equation

$P3=\alpha gh$

where

α=density=1  g/cm^3=1000kg/M^3

H=height=14.1m

g=gravity=3.71m/s^2

solving

P3=(1000)(14.1)(3.71)=52311Pa=52.3kPa

P=P1+P2+P3

P=100kPa+92kPa+52.3kPa=244.3kPa

finally to solve the problem we remember that the pressure is the force exerted on the area

$P=\frac{F}{A} \\F=PA\\F=(244.3kPa)(2m^2)=488.6KN$