Calculating the Net Downward Force on a Cylindrical Water Tank on Mars

AI Thread Summary
To calculate the net downward force on a cylindrical water tank on Mars, the pressure exerted by the water and air must be considered. The pressure at the water's surface is 110 kPa, and the depth of the water is 14.5 m, leading to a downward force from the water of 3.93x10^5 N. The upward force from the outside air is 2.18x10^5 N, and the air pressure inside the tank also contributes to the total downward force. The correct approach involves summing the forces from the water and both air pressures to find the net force. Accurate calculations are essential, particularly when converting units, to ensure the final answer is expressed in kN.
MissKaylaPaige
Messages
33
Reaction score
0

Homework Statement



I have a cylindrical water tank of mars, where acc. due to gravity= 3.71

The pressure at the surface of the water is 110 kPa, and the depth of the water is 14.5 m

The pressure of the air outside the tank is 91 kPa

Find the net downward force on the tanks flat bottom of area 2.4 m^2, exerted by the water and air inside the tank and the air outside the tank



Homework Equations



p= F/A



The Attempt at a Solution



alright, I found the downward force to be 3.93x10^5 N
upward force to be 2.18x10^5

I thought the sum of these forces would give me the net forces but I must've been mistaken because the answer was wrong.
 
Physics news on Phys.org
You've got the water pushing down, so you need to find the volume of water and figure out the mass(I guess you assume 1 mL= 1g) and mg(but not the normal g!)will be the downward force from the water alone

Also the air is pushing down on it, as is the outside air, both can be found using your equation

Don't forget if you use kPa and m^2 you get your answer in kN
 
I know the force from the outside air is 2.18x10^5 N, and the force of the water inside is 3.93x10^5 N, so if I find the force of the air inside the tank as well, the sum of them will give me what I'm looking for?
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...

Similar threads

Back
Top