Moment produced by a fluid of varying density

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Discussion Overview

The discussion centers around a homework problem involving the calculation of the force acting on a gate submerged in a fluid with varying density. The problem requires the application of fluid mechanics principles, particularly in relation to pressure and moments produced by the fluid's weight.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • The original poster presents a detailed attempt at solving for the force acting at the top of the gate, including the setup of the problem and the equations used.
  • Calculations are shown to yield a force of 73.902 kN, which the poster believes is correct based on their work.
  • Another participant checks the calculations and agrees that they appear correct, expressing confusion over the discrepancy with the professor's answer of 63.1 kN.
  • The original poster suggests that the professor may have provided an incorrect answer for a different question and plans to clarify this with the professor.

Areas of Agreement / Disagreement

There is no consensus on the correct answer, as the original poster's calculations yield a different result than that provided by the professor. Participants agree that the calculations check out but remain uncertain about the source of the discrepancy.

Contextual Notes

The discussion does not address potential assumptions or limitations in the setup of the problem or the calculations performed. The absence of a diagram may also affect the clarity of the problem.

wahaj
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Homework Statement


Find F, the force acting at the top of the gate.
The gate is 2m high and 2m into the page. It is hinged at the bottom (red dot in the diagram)
Density of fluid varies linearly. 1000 kg/m3 at the top and 1600 kg/m3 at the bottom (depth of 4m)


Homework Equations



M_s=-\int \int \vec{r} \times \hat{n} \ P dA

The Attempt at a Solution


My origin is at the red dot. z is positive upwards and x is positive towards the left into the fluids
\vec{r} \times \hat{n} = z \hat{k} \times \hat{i} = z \hat{j}
I'm going to ignore the vector part and also the negative sign of this since I only need the magnitude.
P= -\rho g z + C \ (negative\ because\ g\ is\ negative) \\<br /> P = 0\ when\ z = 4 \\<br /> 0= -4 \rho g + C \\ C = 4 \rho g \\<br /> P = \rho g (4 - z) \\<br /> \rho = 1000 \ when\ z = 4 \ \ \ \rho = 1600 \ when\ z = 0 \\<br /> \rho = mz +b \\<br /> 1000 = 4m +b \ \ \ \ \ \ \ \ \ \ \ 1600 = 0m+b \\<br /> \rho = 1600 - 150z \\ <br /> dA = dz dy , \ limits: \ [0,2],[0,2] \\<br /> M = g \int_0^2 \int_0^2 z(1600 - 150z)(4 - z) \ dzdy \\<br /> M = 147.804 kN \\<br /> 147.804kN = 2F \\<br /> F = 73.902 kN
The actual answer is 63.1 kN. What am I doing wrong?
 

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Is there supposed to be a diagram attached?
 
I did attach the diagram. I don't know where it went. Let me try again
Edit: There we go
 
Your calculations check. I can't see how F = 63.1 kN.
 
That's the answer my professor gave. If I am right that means the professor must have given the answer for a different question by mistake. I will ask him about this when classes start again. Thanks for checking my work, I've been losing my mind trying to find the mistake in my work.
 

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