Fluid Hydrostatics: Homework Solutions & Explanations

In summary, the conversation discusses the problem of determining the correct vertical force on a conical surface due to fluid pressure. Two methods were attempted, with the first one not producing the correct result. The second method, which considered the equilibrium of the fluid element surrounding the cone, yielded the correct answer. The participants also question if the first method should have worked and ask for confirmation and explanation.
  • #1
Steve_112
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Homework Statement


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Homework Equations

The Attempt at a Solution


My first confusion is with the pressure measuring device, i just converted the units into Pa by dividing by 1.02x10-5 such that:
Pressure at the top of the tank = (0.07/1.02e-5) - (9800*0.3) = 3922.75Pa
So on the cone the average pressure is (3922.75 + 0.05*9800) = 4412.75Pa
The surface on which this force acts is the top part of the cone, where r = h*tan(30) , r = 0.0577, which means the surface is 0.031m^2
The force acts Vertically and horizontally but the horizontal components cancel each other out (right?) So the vertical force is:
4412.75*0.031*cos(60) = 68.4N, which isn't the correct result.

This is the same method i was using to solve problems related to tilted rectangular plates, however now it is a conical surface should i be solving it differently ?

Any help would be much appreciated, thank you!
 

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  • #2
Well i think i solved it, but i don't know why the first method didn't work!

I just considered the equilibrium of the fluid element surrounding the cone and bounded by the cylinder of radius r=0.0577.
The volume of this fluid element is
Vcyl - Vcone = 1.05e-3 - 3.49e-4 = 7e-4 m^3
Weight of fluid element = 7e-4 * 9800 = 6.83N
The upward force on this fluid element due to pressure is:
(pi*0.0577^2*(3922.75+0.1*9800) = 51.28N
The two horizontal forces cancel each other out thus the vertical force is 51.28 - 6.83N = 44.45N = S
Its the right answer but maybe its just all one big coincidence!
Could anyone confirm if this is correct, and also explain what is wrong with the first method.
Thanks!
 

FAQ: Fluid Hydrostatics: Homework Solutions & Explanations

1. What is fluid hydrostatics?

Fluid hydrostatics is a branch of fluid mechanics that deals with the study of fluids at rest, specifically the principles and laws governing the behavior of fluids under the influence of gravity.

2. What are some examples of applications of fluid hydrostatics?

Some common examples of applications of fluid hydrostatics include the design and construction of dams, bridges, and ships, as well as the study of atmospheric pressure and weather patterns.

3. How is pressure in a fluid measured?

Pressure in a fluid is measured using a device called a manometer, which consists of a U-shaped tube filled with a fluid (such as mercury) and attached to the fluid being studied. The difference in height between the two sides of the manometer is used to calculate the pressure.

4. What is Pascal's law and how does it relate to fluid hydrostatics?

Pascal's law states that the pressure exerted by a fluid in a closed container is transmitted equally and undiminished in all directions. This means that the pressure at any point in a fluid at rest is the same in all directions, and it is a fundamental principle in fluid hydrostatics.

5. How does the density of a fluid affect its hydrostatic pressure?

The density of a fluid directly affects its hydrostatic pressure. This is because the pressure at a certain depth in a fluid is equal to the weight of the fluid column above that point. Therefore, the denser the fluid, the greater the pressure at a given depth.

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