Help solving Fluid Force problem

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Homework Help Overview

The discussion revolves around a fluid mechanics problem involving the calculation of force on a hatch in a tank filled with water. The tank dimensions and water levels are specified, and the participants are attempting to apply integral calculus to determine the force exerted by the water on the hatch.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the bounds of integration for the force calculation, with some confusion about the correct values for 'a' and 'b'. There is also a focus on the definition of variables, particularly how 'y' is being measured in relation to the tank's dimensions.

Discussion Status

There is ongoing clarification regarding the setup of the integral and the definitions of the variables involved. Some participants are questioning the correctness of the integral used and the interpretation of the problem, indicating a productive exploration of the concepts involved.

Contextual Notes

Participants note challenges with integrating the equation correctly and express uncertainty about the bounds and the application of the integral. There is also mention of a typo in the problem statement that has been corrected, which may affect the discussion.

Qbit42
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Homework Statement


A 2.20m wide tank is filled with water totaling 1.87m (H1). There is a hatch cosisting of a portion of a wall of the tank which is 1.15m(H2) in length (measured from the bottom of the tank). What is the Force on the hatch resulting from the water?

Homework Equations


F = INTEGRAL<UPPERBOUND = b, LOWERBOUND = a> ( wgp(b-y)dy ) = 0.5wgp[b^2 - a^2]

p = density of water = 1000 kg/m^3

The Attempt at a Solution


I've tried this problem many different times and I can't seem to get the correct answer. From my point of view the lower bound of integration would be 0.72 (1.87 -1.15) and the upper bound would be 1.87. This gives an answer of 32108 N, which is incorrect. Any help is greatly appreaciated!

Edit: Sorry about being unable to use the integral in the pallete correctly, I can't figure out how to but in bounds (if you can).
 

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Qbit42 said:
1.87m (H1). 1.51m(H2)

[From my point of view the lower bound of integration would be 0.72 (1.87 -1.15) and the upper bound would be 1.87.

Qbit42 said:
F = INTEGRAL<UPPERBOUND = b, LOWERBOUND = a> ( wgp(b-y)dy ) = 0.5wgp[b^2 - a^2]
I'm also not quite sure this integral is fine...
 
Fixed up that typo. The integral is one which i got from my physics textbook. Well actually the integral is just 0.5wpgb^2 since the problem in the text is asking about the pressure over the entire wall (ie from 0 to b) but all i did was make the lower bound non-zero.
 
Last edited:
How are you defining y? And what are you using now for a and b?

The key here, which you seem to be aware of, is that the pressure is p*g*depth.
 
Well a is the point from which I'm measuring the liquid and b is the point at which I stop measuring. So a = 0.72 m and b = 1.87 m. As for how y is defined its the distance from the top (ie a) to a infindesimal strip (ie dA = wdy).
 

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