You are assigned the design of a cylindrical, pressurized water tank for a futur

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 meters per second per second. The pressure at the surface of the water will be 105 kPa , and the depth of the water will be 13.9 m . The pressure of the air in the building outside the tank will be 85.0 kPa .

Find the net downward force on the tank's flat bottom, of area 2.10 m2 , exerted by the water and air inside the tank and the air outside the tank.

Express your answer numerically in Newtons, to three significant figures.

Explanation / Answer

Given

   acceleration due to gravity on Mars is g = 3.71 m/s2,

   pressure at the surface of the water is P = 105*10^3 Pa,

   pressure of air out side the tank is 85 *10^3 Pa

   pressure difference (pressure of environment) is = 20 *10^3 Pa,

we know that the density of water is 1000 kg/m3, and given depth of water si 13.9 m

   from the definition of fluid pressure P_fluid = rho*g*h = 1000*3.71*13.9 = 51569 Pa

now total pressure is sum of pressure of environment and pressure of fluid P = 20000 +51569 = 71569 Pa

area of the surface = 2.10 m2,

force F = P*A = 71569*2.10 = 150294.9 N

The net downward force on the tank's flat bottom, of area 2.10 m2 , exerted by the water and air inside the tank and the air outside the tank is 150294.9 N

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.