Fluid Mechanics Challenging Problem

In summary: Is the diagram something you drew from the problem statement or did it come with the problem statement? If it is your own interpretation then I would say it is wrong, and that the "fluid" is the mercury. But in all other respects your diagram is ok.
  • #1
jahan123
7
0
Homework Statement
how far from the mercury surface must a horizontal line be drawn in a 2 meter square gate whose one side is at the fluid surface so that the hydrostatic forces in the two newly formed surfaces are equal
Relevant Equations
1) F = yAh
where y= specific weight
A is the cross sectional area
h is the distance from the centroid to the surface of the fluid.
Homework Statement: how far from the mercury surface must a horizontal line be drawn in a 2 meter square gate whose one side is at the fluid surface so that the hydrostatic forces in the two newly formed surfaces are equal
Homework Equations: 1) F = yAh
where y= specific weight
A is the cross sectional area
h is the distance from the centroid to the surface of the fluid.

I tried to visualize the problem. Can someone draw the diagram needed. So that i can solve the problem. thanks
 
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  • #2
Welcome to the PF. :smile:
jahan123 said:
Can someone draw the diagram needed. So that i can solve the problem. thanks
Actually, you are the one who needs to draw the diagram. I'm not able to visualize the problem setup anyway, let alone try to draw the diagram.

You can "attach a file" to your next reply to show the setup. Also please label the forces (on the gate I guess?). Thank you.
 
  • #3
berkeman said:
Welcome to the PF. :smile:

Actually, you are the one who needs to draw the diagram. I'm not able to visualize the problem setup anyway, let alone try to draw the diagram.

You can "attach a file" to your next reply to show the setup. Also please label the forces (on the gate I guess?). Thank you.
 

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  • #4
thanks
 
  • #5
That helps enormously, but to be certain, is this a correct description?
A reservoir of mercury is bounded on one side by a 2m square vertical plate. If we notionally divide the plate into upper and lower rectangles, how should we draw the boundary so that the total pressure exerted by the mercury is the same on both rectangles?

If correct, what is the pressure at depth x from the surface of the mercury? What total force does that give for the lower rectangle?
 
  • #6
haruspex said:
That helps enormously, but to be certain, is this a correct description?
A reservoir of mercury is bounded on one side by a 2m square vertical plate. If we notionally divide the plate into upper and lower rectangles, how should we draw the boundary so that the total pressure exerted by the mercury is the same on both rectangles?

If correct, what is the pressure at depth x from the surface of the mercury? What total force does that give for the lower rectangle?
can you please draw a diagram on what you are saying. I can't even visualize what are you saying.
 
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  • #7
haruspex said:
That helps enormously, but to be certain, is this a correct description?
A reservoir of mercury is bounded on one side by a 2m square vertical plate. If we notionally divide the plate into upper and lower rectangles, how should we draw the boundary so that the total pressure exerted by the mercury is the same on both rectangles?

If correct, what is the pressure at depth x from the surface of the mercury? What total force does that give for the lower rectangle?
also can i ask what's the difference between the mercury of the open surface to the fluid surface?
 
  • #8
My a
haruspex said:
That helps enormously, but to be certain, is this a correct description?
A reservoir of mercury is bounded on one side by a 2m square vertical plate. If we notionally divide the plate into upper and lower rectangles, how should we draw the boundary so that the total pressure exerted by the mercury is the same on both rectangles?

If correct, what is the pressure at depth x from the surface of the mercury? What total force does that give for the lower rectangle?
my attempt of my solution

I divided the gate into two rectangles. the base are still the same. i think that the fluid surface must be below the mercury open surface. I let Y be the height of the fluid surface to the open mercury surface. Hence 2-Y be the height of the other rectangle.

I determined the center of gravity of the first rectangle marked as G1, and the center of pressure P1 where the force F1 is acting.

From hydrostatic pressure
F1=wAh
F1= w(2)(y)(y/2)

the same procedure applies to the second rectangle
F2= w(2)(2-y)((2-y)/2)

since F1 and F2 are equal i found out that
y=1m
is that the answer?
 
  • #9
jahan123 said:
whats the difference between the mercury of the open surface to the fluid surface?
Is the diagram something you drew from the problem statement or did it come with the problem statement? If it is your own interpretation then I would say it is wrong, and that the "fluid" is the mercury. But in all other respects your diagram is ok.
jahan123 said:
F1= w(2)(y)(y/2)
Yes, for the top part.
jahan123 said:
the same procedure applies to the second rectangle
F2= w(2)(2-y)((2-y)/2)
No. You have not allowed for the fact that the pressure is higher at all points in the lower rectangle because of the weight above.
One way to calculate it is to find the the total force from the mercury and subtract the force on the top rectangle.
 
  • #10
haruspex said:
Is the diagram something you drew from the problem statement or did it come with the problem statement? If it is your own interpretation then I would say it is wrong, and that the "fluid" is the mercury. But in all other respects your diagram is ok.

Yes, for the top part.

No. You have not allowed for the fact that the pressure is higher at all points in the lower rectangle because of the weight above.
One way to calculate it is to find the the total force from the mercury and subtract the force on the top rectangle.
but the question is where should be the fluid surface be so that the hydrostatic forces are equal.
 
  • #11
jahan123 said:
but the question is where should be the fluid surface be so that the hydrostatic forces are equal.
No, it doesn’t say that. You may be getting confused by two things: mentioning "fluid" as though it is something other than the mercury, but they just mean the mercury; and the use of the word "surface" to refer to two different things: the horizontal surface of the mercury and the vertical surface(s) of the gate.
Here's how I read it:

"how far from the mercury surface must a horizontal line be drawn in (across) a (vertical) 2 meter square gate whose one side is at (i.e. reaches up to) the (mercury) fluid surface so that the hydrostatic forces (from the mercury) in (on) the two newly formed (in an imaginary sense, by having drawn the line across) (vertical) surfaces (i.e. the parts of the gate above and below this line) are equal.
 

1. What is fluid mechanics and why is it important?

Fluid mechanics is the study of how fluids (liquids and gases) behave and interact with their surroundings. It is important because it helps us understand and predict the motion of fluids, which is crucial in many engineering and scientific applications such as designing airplanes and predicting weather patterns.

2. What are some common challenging problems in fluid mechanics?

Some common challenging problems in fluid mechanics include turbulent flow, boundary layer separation, and multiphase flow. These problems can be complex and difficult to solve due to the nonlinear nature of fluid behavior.

3. How do scientists and engineers approach challenging problems in fluid mechanics?

Scientists and engineers use mathematical equations and computational models to study and analyze fluid mechanics problems. They also conduct experiments and gather data to validate their theories and models.

4. Can you provide an example of a challenging problem in fluid mechanics?

One example of a challenging problem in fluid mechanics is the study of turbulent flow in a pipe. Turbulent flow is characterized by chaotic and unpredictable motion, making it difficult to model and understand. This problem has real-world applications in industries such as oil and gas, where efficient transportation of fluids through pipelines is crucial.

5. How does fluid mechanics relate to other fields of science and engineering?

Fluid mechanics has applications in many other fields of science and engineering, such as aerospace engineering, chemical engineering, and meteorology. Understanding fluid behavior is essential in designing and optimizing systems and processes in these fields.

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